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3 


ROBINSON'S 


PROGRESSIVE 


Practical  Aeithmetic; 


CONTAININa 

THE     THEORY     OF     NUMBERS,     IN      CONNECTION     WITH     CONCISE 

ANALYTIC    AND    SYNTHETIC    METHODS    OF    SOLUTION,    AND 

DESIGNED  AS  A  COMPLETE  TEXT-BOOK  ON  THIS  SCIENCE, 


FOR 


COMMON    SCHOOLS   AND   ACADEMIES. 


BY 

DANIEL  W.  FISH,   A.M., 

AUTHOR    OF    THE    TABLB-BOOK,     PRIMARY    ANO     INTELLECTUAL    ARITHMETICS, 
1CUL.IMENTS,    AND    THE    "SHORTER    COURSE." 


IVISON,    BLAKEMAN,    TAYLOR    &    CO., 

NEW    YORK    AND    CHICAGO. 
1881. 


mmAnon  ube. 

ROBINSON'S 

Mathematical    Series. 

traded  to  the  wants  of  Primary,  Intermediate,  Grammar, 
Normal,  and  High  Schools,  Academies,  and  CollegeB, 


•-♦■« 

FrogressiTe  Table  Book. 
Progressire  Primary  Arithmetic. 
Progressire  Intellectual  Arithmetic. 
Badiments  of  Written  Arithmetic. 
JUNIOB-CLASS  ARITHMETIC,  Oral  and  Written.    KEW. 
ProgrestiTe  Practical  Arithmetic. 
Key  to  Practical  Arithmetic. 
ProgresgiTC  Higher  Arithmetic. 
Eey  to  Higher  Arithmetic. 
New  Elementary  Algebra. 
Key  to  New  Elementary  Algebra. 
New  Uniyersity  Algebra. 
Key  to  New  Unirerslty  Algebra. 
New  Geometry  and  Trigonometry.    In  one  vol. 
Qeometry,  Plane  and  Solid.    In  separate  vol. 
Trigonometry,  Plane  and  Spherical.    In  separate  voL 
New  Analytical  Geometry  and  Conic  Sections. 
New  Surreying  and  Narigation. 
New  Differential  and  Integral  Calcolns. 
UniTorsity  Astronomy— Descriptive  and  Physical. 
Key  to  Geometry  and  Trigonometry,  Analytical  Geometry  uid  Conic  Seo> 
tiong,  Surveying  and  NaTigatlon. 

Entered,  according  to  Act  of  Congress,  in  the  year  1858,  and  again  in  the  year  186S,  by 

DANIEL    W.   FISH,    A.M., 

la  }he  Clerk'a  Office  of  the  District  Court  of  the  United  States,  for  the  Northern 
District  of  New  York. 

Co^right^  X877,  by  DauUl  W,  Fuk» 


Add  to  Lib. 
GIFT 


PREFACE. 


"PROGRESS  and  improvement  characterize  almost  every  art  and 
-"-  science  ;  and  within  the  last  few  years  the  science  of  Arithmetic 
has  received  many  imjportant  additions  and  improvements,  which 
have  appeared  from  time  to  time  successively  in  the  different  treatises 
published  upon  this  subject. 

In  the  preparation  of  this  work  it  has  been  the  author's  aim  to 
combine,  and  to  present  in  one  harmonious  whole,  all  these  modem 
improvements,  as  well  as  to  introduce  some  new  methods  and  prac- 
tical operations  not  found  in  other  works  of  the  same  grade  ;  in  short, 
to  present  the  subject  of  Arithmetic  to  the  pupil  more  as  a  science 
than  an  art ;  to  teach  him  methods  of  thought,  and  how  to  reason, 
rather  than  whM  to  do;  to  give  unity,  system,  and  practical  utility  to 
the  science  and  art  of  computation. 

The  author  believes  that  both  teacher  and  pupil  should  have  the 
privilege,  as  well  as  the  benefit,  of  performing  at  least  a  part  of  the 
thinking  and  the  labor  necessary  to  the  study  of  Arithmetic ;  hence 
the  present  work  has  not  been  encumbered  with  the  multiplicity  of 
**  notes,"  '*  suggestions,"  and  superfluous  operations  so  common  to 
most  Practical  Arithmetics  of  the  present  day,  and  which  prevent  the 
cultivation  of  that  self-reliance,  that  clearness  of  thought,  and  that 
vigor  of  intellect,  which  always  characterize  the  truly  educated  mind. 

The  author  claims  for  this  treatise  improvement  upon,  if  not  supe. 
riority  over,  others  of  the  kind  in  the  following  particulars,  viz. :  In 
the  mechanical  and  typographical  style  of  the  work ;  the  open  and 
attractive  page  ;  the  progremve  and  scientific  arrangement  of  the 
subjects;  clearness  and  conciseness  of  definitions;  fullness  and  accu^ 
racy  in  the  new  and  improved  methods  of  operations  and  analyses ; 
brevity  and  perspicuity  of  rvles ;  and  in  th^  very  large  number  of 
examples  prepared  and  arranged  with  special  reference  to  thdr  prac- 
tical utility,  and  their  adaptation  to  the  real  business  of  active  life.    Tlip 

917 


IV  PREFACE. 

answers  to  a  part  of  the  examples  have  been  omitted,  that  the  learner 
may  acquire  the  discipline  resulting  from  verifying  the  operations. 

Particular  attention  is  invited  to  improvements  in  the  subjects  of 
Common  Divisors,  Multiples,  Fractions,  Percentage,  Interest,  Pro- 
portion, Analysis,  Alligation,  and  the  Roots,  as  it  is  believed  these 
articles  contain  some  practical  features  not  common  to  other  authors 
upon  these  subjects. 

The  improvements  in  Percentage  made  necessary  by  the  financial 
changes  of  the  last  few  years  are  especially  noticeable.  The  different 
kinds  of  United  States  Securities,  Bonds,  and  Treasury  Notes  are 
described,  and  their  comparative  value  in  commercial  transactions 
illustrated  by  practical  examples.  The  difference  between  Gold  and 
Currency,  and  the  corresponding  difference  in  prices,  exhibited  in 
trade,  are  taught  and  illustrated,  and  many  other  things  that  every 
commercial  student  and  business  man  ought  to  know  and  understand. 

It  is  not  claimed  that  this  is  a  perfect  work,  for  perfection  is  im- 
possible ;  but  no  effort  has  been  spared  to  present  a  clear,  scientific, 
comprehensive,  and  complete  system,  sufficiently  full  for  the  busi- 
ness man  and  the  scholar  ;  not  encumbered  with  unnecessary  theories, 
and  yet  combining  and  systematizing  real  improvements  of  a  practical 
ind  useful  nature.  How  nearly  this  end  has  been  attained  the  intel- 
Igent  and  experienced  teacher  and  educator  must  determine. 

The  Author. 


NOTICE. 

This  book  has  been  newly  electrotyped  in  the  latest  and  best  style 
of  typography.  No  change  has  been  made  in  the  text,  except  to  cor- 
rect positive  errors,  and  drop  a  few  obsolete  terms  from  the  tables. 

These  improvements  are  designed  to  give  new  life  to  a  book  that 
has  proved  its  real  merits  by  the  very  large  circulation  it  has  obtained. 

Bbookltn,  Jviyt  18TZ. 


CONTENTS. 


SIMPLE    NUMBERS. 


PAGE 

DEFunriONS T 

Roman  Notation 8 

Table  of  Roman  Notation  .  9 

Arabic  Notation 10 

Numeration  Table 15 

Laws  and  Rules  for  Notation  and 

Numeration 17 

Addition 20 

Subtraction 28 

Multiplication 35 

Contractions 42 


PAGS 

Division 47 

Contractions.. 56 

Applications  of  Preceding  PtULES.  60 

General  Principles  op  Division  ...  64 

Exact  Divisors 66 

Prime  Numbers 67 

Factoring  Numbers 67 

Cancellation 69 

Greatest  Common  Divison 73 

Multiples 79 

Classification  op  Numbebs 84 


COMMON    FRACTIONS. 


Definitions,  etc 

General  Principles  op  Fractions. 
Reduction  op  Fractions 


Addition  of  Fractions 96    Promiscuous  Examples 


Subtraction  op  Fractions 98 

Multiplication  op  Fractions 101 

Division  OF  Fractions 106 


112 


DECIMALS. 


Decimal  Notation  and   Numera- 
tion   116 

Reduction  of  Decimals 121 

Addition  op  Declmals 124 


Subtraction  op  Decimals 126 

Multiplication  of  Decimals 127 

Division  op  Decimals 128 


DECIMAL    CURRENCY. 


Notation    and    Numeration    op 

Decimal  Currency.  . 131 

Reduction  of  Decimal  Currency.  132 
Addition  of  Decimal  Currency.  . .  134 
Subtraction  of  Decimal  Currency  135 
Multiplication  op  Decimal  Cur- 
rency   136 

Division  of  Decimal  Currency..  . .  137 
Additional  Applications 139 


When  the  Price   is  an  Auqttot 

Part  OF  A  Dollar 139 

To  Find  the  Cost  op  a  Quantity.  .  140 

To  Find  the  Price  of  One 141 

To  Find  the  Quantity  141 

Articles  Sold  by  the  100  or  1000..  142 

Articles  Sold  by  the  Ton 143 

Bills 144 

Promiscuous  Examples 146 

(v) 


VI 


CONTENTS. 


COMPOUND    NUMBERS. 


PAGE 

Redttction 150 

Definitions,  BTC 150 

English  Money 151 

Trot  Weight 153 

Apothecaries'  Weight 154 

Avoirdupois  Weight 155 

Long  Measure 158 

StTRTEYORS*  LONG   MeASITBB 160 

Square  Measure 161 

Surveyors'  Square  Measure 164 

Cubic  Measure 165 

Liquid  Measure 167 

Dry  Measure 168 

Time 170 

Circular  Measure 172 

Counting  ;  Paper  ;  Books,  etc 173 


PAQH 

Reduction  of  Denominate  Frac- 
tions  175 

Addition  op  Compound  Numbers.  .  182 
Addition   op   Denominate    Frac- 
tions   185 

Subtraction 186 

To  Find  the  Difference  in  Dates  188 

Table 189 

Subtraction  of  Denominate  Frac- 
tions   190 

Multiplication  of  Compound  Num- 
bers   191 

Division 193 

Longitude  AND  Time 195 

Duodecimals 198 

Promiscuous  Examples 208 


PERCENTAGE. 


Definttions,  etc 205 

Commission  and  Brokerage 212 

Stock- Jobbing 216 

United  States  Securities 220 

Gold  Investments 225 

Profit  and  Loss . .  227 

Insurance 233 

Taxes 234 

CusTOM-HousE  Business 237 


Simple  Interest 240 

Partial    Payments    or   Indorse- 
ments  247 

Problems  in  Interest 258 

Compound  Interest 256 

Discount 259 

Banking 262 

Exchange 266 

Equation  of  Payments 271 


RATIO    AND    PROPORTION. 


Ratio 279 

Proportion 282 

Simple  Proportion 283 

Compound  Proportion 289 

Partnership 294 


Analysis 298 

Alligation  Medial 807 

Alligation  Alternate 808 


I  Involution 813 

Evolution 314 

Square  Root 815 

Cube  Root 822 


Arithmetical  Progression. 
Geometrical  Progression.  . 


831 


Promiscuous  Examples 884 

Mensuration 842 

The  Metric  System 845 


PEACTICAL  AEITHMETIC. 


DEFmiTIOl^S.        * 

1.  Quantity  is  any  thing  that  can  be  increased,  dimin* 
ished,  or  measured. 

2.  Mathematics  is  the  science  of  quantity. 

3.  A  Unit  is  one,  or  a  single  thing. 

4.  A  Number  is  a  unit,  or  a  collection  of  units, 

5.  An  Integer  is  a  whole  number. 

6.  The  Unit  of  a  Number  is  one  of  the  collection  of 
units  forming  the  number.  Thus,  the  unit  of  23  is  1 ;  of 
23  dollars,  1  dollar  ;  of  23  feet,  1  foot 

7.  Like  Numbers  are  numbers  that  have  the  same  kind 
of  unit.  Thus,  74, 16,  and  250  ;  7  dollars  and  62  dollars  ; 
4  feet  6  inches,  and  17  feet  9  inches. 

8.  An  Abstract  Number  is  a  number  used  without 
reference  to  any  particular  thing  or  quantity.  Thus,  17  ; 
365;  8540. 

9.  A  Concrete  Number  is  a  number  used  with  refer- 
ence to  some  particular  thing  or  quantity.  Thus,  17  dollars ; 
365  days  ;  8540  men. 

1.  The  unit  of  an  abstract  number  is  1,  and  is  called  Unity. 
a.  Concrete  numbers  are,  by  some,  called  DenomincUe  members.    DenondnaUon 
means  the  name  of  the  unit  of  a  concrete  number. 

10.  Arithmetic  is  the  Science  of  numbers,  and  the  Art 
of  computation. 

11.  A  Sign  is  a  character  used  to  indicate  an  operation, 
or  for  abbreviating  an  expression. 

13.  A  Rule  is  a  prescribed  method  for  performing  an 
operation. 

Define  quantity.    Mathematics.     A  unit.     A  number.  An  integer. 

The  unit  of  a  number.     Like  numbers.     An  abstract  number.     A 

concrete  number.     The  unit  of  an  abstract  number.  Denominate 
numbers.     Arithmetic.     A  sign,  or  symbol.    A  rule. 


SIMPLE    NUMBEES. 


H"OTATION  AIsTD  NUMERATION. 

13.  Notation  is  a  method  of  writing  or  expressing  num- 
bers by  characters. 

14.  Numeration  is  a  method  of  readitig  numbers  ex 
pressed  by  characters. 

15.  Two  systems  of  notation  are  in  general  use — ^th6 

Roman  and  the  AraMc. 

The  Roman  Notation  is  Bnpposed  to  have  been  firet  used  by  the  Bomans ;  hence 
its  name.  The  Arabic  Notation  was  introduced  into  Enrope  by  the  Arabs,  by  whom 
it  was  supposed  to  have  been  invented.  But  investigations  have  shown  that  it  was 
adopted  by  them  about  600  years  ago,  and  that  it  has  been  in  use  among  the  Hin- 
doos more  than  2000  years.  From  this  latter  fact  it  is  sometimes  called  the  Indian 
Notation. 

The  Romak  NoTATi02<r. 

16.  Employs  seven  capital  letters  to  express  numbers. 
Thus, 

Letters.       I  V  X  L  C  D  M 

Values.      one.        Five.        Ten.        Fifty.       ^Sed.  hSd^d.  thJSnd. 

17.  The  Roman  Notation  is  founded  upon  fL^e  principles, 
as  follows : 

1st  Repeating  a  lei  tor  repeats  its  yalue.  Thus,  II  repre- 
sents two,  XX  twenty,  CCC  three  hundred. 

2d.  If  a  letter  of  any  value  is  placed  after  one  of  greater 
value,  its  value  is  to  be  added  to  that  of  the  greater.  Thus. 
XI  represents  eleven,  LX  sixty,  DC  six  hundred. 

3d.  If  a  letter  of  any  value  be  placed  lefore  one  of  greater 
value,  its  value  is  to  be  taken  from  that  of  the  greater. 
Thus,  IX  represents  nine,  XL  forty,  CD  four  hundred. 

Define  notation.  Numeration.  What  systems  of  notation  are  now 
in  general  use?  From  what  are  their  names  derived  ?  Wliat  are  used 
to  express  numbers  in  the  Roman  notation  ?  What  is  the  value  of 
each  ?    What  is  the  first  principle  of  combination ?    Second  ?    Third? 


NOTATION    AND    NUMEEATION. 


9 


4th.  If  a  letter  of  any  value  be  placed  between  two  letters, 
each  of  greater  value,  its  value  is  to  be  taken  from  the  su7n 
of  the  other  two.  Thus,  XIV  represents  fourteen,  XXIX 
twenty-nine,  XCIV  ninety-four. 

5th.  A  bar  or  dash  placed  over  a  letter  increases  its  value 
one  thousand  times.  Thus,  V  signifies  five,  and  V  five  thou- 
sand ;  L  fifty,  and  L  fifty  thousand. 

.Table  of  Eoman  Notation. 


I  =  One. 

XX  =  Twenty. 

n  «  Two. 

XXI  "  Twenty-one. 

Ill  ^  Three. 

XXX  «  Thirty. 

IV  "  Four. 

XL  "  Forty. 

V  "  Five. 

L  "  Fifty. 

VI  "  Six 

LX  "  Sixty. 

VII  "  Seven. 

T.XX  "  Seventy. 

VIII  "  Eight. 

LXXX  "  Eighty. 

IX  «  Nine. 

XC  "  Ninety. 

X  "  Ten. 

C  "  One  hundred. 

IX  «  Eleven. 

CO  "  Two  hundred. 

XII  "  Twelve. 

D  "  Five  hundred. 

XIII  "  Thirteen. 

DC  "  Six  hundred. 

XIV  "  Fourteen. 

M  **  One  thouf^nd. 

XV  «  Fifteen. 

MC  "  One  thousand  one  hundred. 

XVi  "  Sixteen. 

MM  "  Two  thousand. 

XVn  "  Seventeen. 

X  "  Ten  thouBand. 

SLVlii  «  Eighteen. 

C  "  One  hundred  thousand. 

XIX  •*  Nineteen. 

M  "  One  million. 

The  STstem  of  Roman  notation  is  not  well  adapted  to  the  purposes  of  numerical 
calculation  ;  it  is  principally  confined  to  tiie  numbering  of  chapters  and  sections  of 
books,  public  documents,  etc 

Express  the  following  numbers  by  letters : 

1.  Eleven.  Ans,  XL 

2.  Fifteen.  .  Ans. 


.  Fourth?  Fifth?  Repeat  the  table.  What  is  the  value  of  LVII ? 
CLXXm?  XCVIII?  CDXXXII?  XCIX?  DCXIX?  "VMDCCXLIX? 
MDXXVCDLXXXIX?  To  what  uses  is  the  Roman  notation  now 
principally  confined  1 


10  SIMPLE    iqUMBEES. 

3.  Twenty-five. 

4.  Thirty-nine. 

5.  Forty-eight. 

6.  Seventy-seven. 

7.  One  hundred  fifty-nine. 

8.  Five  hundred  ninety-four* 

9.  One  thousand  five  hundred  thirty-eight. 

10.  One  thousand  nine  hundred  ten. 

11.  Express  the  present  year. 

The  Akabic  Notatiok. 

18.  Employs  ten  characters  or  figures  to  express  num^ 
bers.    Thus, 

Figures.    0123        456789 

Karnes,  Cipher.   One.  Two.  Three.  Four.  Five.    Six.    Seven.  Eight.  Nine. 

19.  The  first  character,  or  cipher,  is  called  naught,  be- 
cause it  has  no  value  of  its  own.  The  other  nine  characters 
are  called  significant  figures,  because  each  has  a  value  of  its 
own. 

20.  The  significant  figures  are  also  called  Digits,  a  word 
derived  from  the  Latin  term  digitus,  which  signifies /w^er. 

21.  The  naught  or  cipher  is  also  called  nothing,  and  zero. 

The  ten  Arabic  characters  are  the  Alphabet  of  Arithmetic, 
and  by  combining  them  according  to  certain  principles,  all 
numbers  can  be  expressed.  We  will  now  examine  the  most 
impoi-tant  of  these  principles.* 

22.  Each  of  the  nine  digits  has  a  value  of  its  own; 
hence  any  number  not  greater  than  9  can  be  expressed  by 
one  figure. 

*  Fractional  and  decimal  notation,  and  the  notation  of  componnd  numbers,  -will 
be  discoBsed  in  their  appropriatai  places. 

What  are  used  to  express  numbers  in  the  Arabic  notation  ?  What 
is  the  value  of  each  ?  What  general  name  is  given  to  the  significant 
figures  ?    Why  ?    Numbers  less  than  ten,  how  expressed  ? 


NOTATION"    AND     NUMEKATION.  11 

33.  As  there  is  no  single  character  to  represent  ten,  we 
express  it  by  writing  the  unit,  1,  at  the  left  of  the  cipher,  0, 
thus,  10.    In  the  same  manner  we  represent 


2  tens, 

3  tens, 

4  tens. 

5  tens. 

6  tens. 

7  tens, 

8  tens. 

9  tens. 

or 

or 

or 

or 

or 

or 

or 

or 

Twenty, 

Thirty, 

Forty, 

Fifty, 

Sixty, 

Seventy, 

Eighty, 

Ninety, 

20; 

30; 

40; 

50; 

60; 

TO; 

80; 

90. 

24,  When  a  number  is  expressed  by  two  figures,  the  right 
hand  figure  is  called  units,  and  the  left  hand  figure  tens. 

We  express  the  numbers  between  10  and  20  by  writing 
the  1  in  the  place  of  tens,  with  each  of  the  digits  respec- 
tively  in  the  place  of  units.     Thus, 


Eleven, 

Twelve, 

Thirteen, 

Fourteen, 

Fifteen, 

Sixteen, 

Seventeen, 

Eighteen, 

Nineteen, 

n, 

13, 

13, 

14, 

15, 

16, 

17, 

18, 

19. 

In  like  manner  we  express  the  numbers  between  20  and 
30,  between  30  and  40,  and  between  any  tAVO  successive  tens. 
Thus,  21,  22,  23,  24,  25,  26,  27,  28,  29,  34,  47,  56,  72,  93. 
The  greatest  number  that  can  be  expressed  by  two  figures 
is  99. 

25.  We  express  one  hundred  by  writing  the  unit,  1,  at 
the  left  hand  of  two  ciphers,  or  the  number  10  at  the  left 
hand  of  one  cipher  ;  thus,  100.  In  like  manner  we  write 
two  hundred,  three  hundred,  etc.,  to  nine  hundred.     Thus, 

One        Two        Three       Four        Five  Six         Seven       Eight        Nine 

hundred       hundred,      hundred,      hundred,      hundred,       hundred,       hundred,      hundred,      hundred, 

100,     200,      300,     400,     500,     600,      700,      800,     900. 

26.  When  a  number  is  expressed  by  three  figures,  the 
right  hand  figure  is  called  units,  the  second  figure,  tens,  and 
the  left  hand  figure,  hundreds. 

As  the  ciphers  have,  of  themselves,  no  value,  but  are 
always  used  to  denote  the  absence  of  value  in  the  places  they 

Tens,  liow  expressed  ?  The  right  hand  figure  called,  what  ?  I-.eft 
hand  figure,  what  ?  What  is  the  greatest  numher  that  can  be  expressed 
hy  two  figures  ?  One  hundred,  how  expressed  ?  When  numbers  are 
expressed  by  three  figures,  what  names  are  given  to  each  ? 


12  SIMPLE    1?"UMBERS. 

occupy,  we  express  tens  and  units  with  hundreds,  by  writ- 
ing, in  place  of  the  ciphers,  the  numbers  representing  the 
tens  and  units.  To  express  one  hundred  fifty  we  write  1 
hundred,  5  tens,  and  0  units ;  thus,  150.  To  express  seven 
hundred  ninety-two,  we  write  7  hundreds,  9  tens,  and  2 
units;  thus, 

M  ^  S 

w      e«      t> 
7      9      3 

The  greatest  number  that  can  be  expressed  by  three  figures 

is  999. 

Examples  for  Practice. 

1.  Write  one  hundred  twenty-five. 

2.  Write  four  hundred  eighty-three. 

3.  Write  seven  hundred  sixteen. 

4.  Express  by  figures  nine  hundred. 

6.  Express  by  figures  two  hundred  ninety. 

6.  Write  eight  hundred  nine. 

7.  Write  five  hundred  five. 

8.  Write  five  hundred  fifty-seven. 

27.  We  express  one  thousand  by  writing  the  unit,  1,  at 
the  left  hand  of  three  ciphers,  the  number  10  at  the  left 
hand  of  two  ciphers,  or  the  number  100  at  the  left  hand  of 
one  cipher  ;  thus,  1000.  In  the  same  manner  we  write  tAvo 
thousand,  three  thousand,  etc.,  to  nine  thousand ;  thus. 

One         Two       Three       Four        Five         Six         Seven      Eight        Nine 

flioaiand,     ihoiuand,     thoiuand,     thoosaud,      thousand,     thousand,      thousand,     thousand,     thousand. 

1000,   2000,  3000,    4000,    5000,   6000,    7000,  8000,    9000. 

28.  When  a  number  is  expressed  by  four  figures,  the 
places,  commencing  at  the  right  hand,  are  unitSf  tens,  hun- 
dreds, thousands. 

Use  of  the  cipher,  what  ?  Greatest  number  that  can  be  expressed  by 
three  figures  V  One  thousand,  how  expressed  ?  How  many  figures 
Used  1    J^ames  of  eaob  'i 


NOTATION    AND    NUMEEATION.  13 

To  express  hundreds,  tens,  and  units  with  thousands,  we 
write  in  each  place  the  figure  indicating  the  number  we  wish 
to  express  in  that  place.  To  write  four  thousand  two  hun- 
dred sixty-nine,  we  write  4  in  the  place  of  thousands,  2  in 
the  place  of  hundreds,  6  in  the  place  of  tens,  and  9  in  the 
place  of  units  ;  thus, 

00  . 

73  OD 

i  1 
i  s  i  n 

4      2      6      9 

The  greatest  number  that  can  be  expressed  hy  four  figures 
is  9999. 

Examples  for  Pbactice. 

Express  the  following  numbers  by  figures  : 

1.  One  thousand  two  hundred. 

2.  Five  thousand  one  hundred  sixty. 

3.  Three  thousand  seven  hundred  forty-one. 

4.  Eight  thousand  fifty-six. 

5.  Two  thousand  ninety. 

6.  Seven  thousand  nine. 

7.  One  thousand  one. 

8".  Nine  thousand  four  hundred  twenty-seven. 

9.  Four  thousand  thirty-five. 
10.  One  thousand  nine  hundred  four. 
Eead  the  following  numbers  : 

IL         76  ;     128  ;     405  ;     910  ;     116  ;  3416  ;  1025. 
12.     2100;  5047;  7009;  4670;  3997;  1001. 

39.  Next  to  thousands  come  iiens  of  thousands,  and  next 
to  these  come  hundreds  of  thousands,  as  tens  and  hundreds 
Come  in  their  order  after  units.  Ten  thousand  is  expressed 
by  removing  the  unit,  1,  one  place  to  the  left  of  the  place 

Greatest  number  expressed  by  four  figures*^  Tens  of  thousands, 
Low  expressed  ?    Hundred,-,  of  thousands  ? 


14  SIMPLE    NUMBERS. 

of  thousands,  or  by  writing  it  at  the  left  hand  of  fotu 
ciphers ;  thus,  10000 ;  and  one  hundred  thousand  is  ex- 
pressed by  removing  the  unit,  1,  still  one  place  further  to 
the  left,  or  by  writing  it  at  the  left  hand  of  five  ciphers ; 
thus,  100000.  We  can  express  thousands,  tens  of  thou- 
sands, and  hundreds  of  thousands  in  one  number,  in  the 
same  manner  as  we  express  units,  tens,  and  hundreds  in  one 
number.  To  express  five  hundred  twenty-one  thousand 
eight  hundred  three,  we  write  5  in  the  sixth  place,  counting 
from  units,  2  in  the  fifth  place,  1  in  the  fourth  place,  8  in 
the  third  place,  0  in  the  second  place  (because  there  are  no 
tens),  and  3  in  the  place  of  units  ;  thus. 


^a 

o1 

a 

V. 

'O  a 

o  "^ 

d 

§^ 

§  3 
H  2 

^ 

o 

W*" 

-B 

B* 

W 

5 

2 

1 

8 

Eh  P 

0      3 

The  greatest  number  that  can  be  expressed  by ^i;e  figures 
ifi  99999 ;  and  by  six  figures,  999999. 

Examples  for  Practice. 

Write  the  following  numbers  in  figures  : 

1.  Twenty  thousand. 

2.  Forty-seven  thousand. 

3.  Eighteen  thousand  one  hundred. 

4.  Twelve  thousand  three  hundred  fifty. 

5.  Thirty-nine  thousand  five  hundred  twenty-two. 

6.  Fifteen  thousand  two  hundred  six. 

7.  Eleven  thousand  twenty-four. 

8.  Forty  thousand  ten. 

9.  Sixty  thousand  six  hundred. 

10.  Two  hundred  twenty  thousand. 

11.  One  hundred  fifty-six  thousand. 

12.  Eight  hundred  forty  thousand  three  hundred. 


Greatest  number  expressed  by  five  figures  ?    Six  fi^fures  ? 


NOTATION     AND    NUMEEATION 


15 


13.  Five  hundred  one  thousand  nine  hundred  sixty-four. 

14.  One  hundred  thousand  one  hundred. 

15.  Three  hundred    thirteen  thousand    three  hundred 
thirteen. 

16.  Seven  hundred  eighteen  thousand  four. 

17.  One  hundred  thousand  ten. 
Eead  the  following  numbers  : 


18.  5006 

19.  36741 

20.  200200 


12304 


96071 ; 
13061 ; 
90402  : 


5470 


49000 ; 
218094 : 


203410. 
100010. 
100101, 


400560 ; 
75620  ; 

For  convenience  in  reading  large  numbers,  point  them  off, 
by  commas,  into  periods  of  three  figures  each,  counting  from 
the  right  hand  or  unit  figure.  This  pointing  enables  us  to 
read  the  hundreds,  tens,  and  units  in  each  period  with 
facility. 

30.  "Next  above  hundreds  of  thousands  we  have,  succes- 
sively, units,  tens,  and  hundreds  of  millions,  and  then  fol- 
low units,  tens,  and  hundreds  of  each  higher  name,  as  seen 
in  the  following : 

Numeration  Table. 


Periods. 

6th. 

5th. 

4tli. 

3d. 

2d. 

Isfc. 

Name.       < 

i 

QQ 

§ 

i 

i 

•  Quad; 

! 

1 

5 

1 

Orders 

OF 

Units. 

undreds. 

ens. 

nits. 

undreds.  ] 

ens. 

nits.          1 

undreds. 

ens. 

nits. 

undreds.  ] 

ens. 

nits.         ) 

undreds.  j 

ens. 

nits. 

undreds. 

3ns. 

nits. 

I.  W^t) 

WEHt) 

tXlEnt) 

WHti 

WEh|:d 

WH^ 

Number. 

5  3  0, 

0  4  5, 

3  7  0, 

0  3  6, 

4  0  8, 

0  6  0. 

The  nunc 

iber  is  read  530  quadrillion, 

46  trillion,  370  billion,  36 

million,  408 

thousand  60. 

How  may  figures  be  pointed  off?  One  million,  how  expressed? 
Next  period  above  millions,  what  ?  Give  the  name  of  each  successive 
period. 


16  SIMPLE    NUMBERS. 

This  is  called  the  French  method  of  pointing  off  the  periods,  and  Is  the  one  in 
general  use  in  this  country. 

31.  Figures  occupying  different  places  in  a  number,  as 
units,  tens,  hundreds,  etc.,  are  said  to  express  different 
orders  of  units. 

Simple  units  are  called  units  of  the  first    order. 

Tens  ''      "       ''      ''   ''  second    " 

Hundreds  "      "       ''      ''   ''   third      " 

Thousands  "      "       "       "   ''  fourth    " 

Tens  of  thousands  "      ''       "       "   ''  fifth       " 

and  so  on.  Thus,  452  contains  4  units  of  the  third  order, 
5  units  of  the  second  order,  and  2  units  of  the  first  order. 
1,030,600  contains  1  unit  of  the  seventh  order  (millions),  3 
units  of  the  fifth  order  (tens  of  thousands),  and  6  units  of 
the  third  order  (hundreds). 

Examples  for  Practice. 

"Write  in  figures,  and  read  the  following  numbers  : 

1.  One  unit  of  the  third  order,  four  of  the  second. 

2.  Three  units  of  the  fifth  order,  two  of  the  third,  one  ol 
the  first. 

3.  Eight  units  of  the  fourth  order,  five  of  the  second. 

4.  Two  units  of  the  seventh  order,  nine  of  the  sixth,  four 
of  the  third,  one  of  the  second,  seven  of  the  first. 

5.  Three  units  of  the  sixth  order,  four  of  the  second. 

6.  Nine  units  of  the  eighth  order,  six  of  the  seventh,  three 
of  the  fifth,  seven  of  the  fourth,  nine  of  the  first. 

7.  Four  units  of  the  tenth  order,  six  of  the  eighth,  four 
of  the  seventh,  two  of  the  sixth,  one  of  the  third,  five  of  the 
second. 

8.  Eight  units  of  the  twelfth  order,  four  of  the  eleventh, 
six  of  the  tenth,  nine  of  the  seventh,  three  of  the  sixth,  five 
of  the  fifth,  two  of  the  third,  eight  of  the  first. 

Units  of  diflferent  orders  are  what  ? 


NOTATION    AND    NUMEKATION.  17 

33.  From  the  foregoing  explanations  and  illustrations, 
several  important  principles  are  derived,  which  we  will  now 
present. 

1st.  Figures  have  two  values,  Simple  and  Local. 

The  Simple  Value  of  a  figure  is  its  value  when  taken 
alone ;  thus,  2,  5,  8. 

The  Liocal  Value  of  a  figure  is  its  value  when  used  with 
another  figure  or  figures  in  the  same  number  ;  thus,  in  842 
the  simple  values  of  the  several  figures  are  8,  4,  and  2  ;  but 
the  local  value  of  the  8  is  800  ;  of  the  4  is  4  tens,  or  40 ; 
and  of  the  2  is  2  units. 

When  a  figure  occupies  units'  place,  its  simple  and  local  values  are  the  same, 

2d.  A  digit  or  figure,  if  used  in  the  second  place,  ex- 
presses tens ;  in  the  third  place,  hundreds ;  in  the  fourth 
place,  thousands  ;  and  so  on. 

3d.  As  10  units  make  1  ten,  10  tens  1  hundred,  10  hun- 
dreds 1  thousand,  and  10  units  of  any  order,  or  in  any  place, 
make  one  unit  of  the  next  higher  order,  or  in  the  next 
place  at  the  left,  we  readily  see  that  the  Arabic  method  of 
notation  is  based  upon  the  following 

Two  General  Laws, 

L  Tlie  value  of  units  of  the  different  orders  increases  from 
right  to  left,  and  decreases  from  left  to  right,  in  a  tenfold 
ratio, 

II.  Every  removal  of  a  figure  one  place  to  the  left,  increases 
its  heal  value  tenfold;  and  every  removal  of  a  figure  oneplac^ 
io  the  right  diminishes  its  local  value  tenfold, 

6  is    6  units. 
60  is  10  times  6  units. 
600  is  10  times  6  tens. 
6000  is  10  times  6  hundreds. 
60000  is  10  times  6  thousands. 

First  principle  derived  ?  What  is  the  simple  value  of  a  fi^re  ?  Local? 
Second  principle  ?    Third  ?    First  law  of  Arabic  notation  ?    Second  ? 


18  SIMPLE    NUMBERS. 

4th.  The  local  value  of  a  figure  depends  upon  its  place 
from  units  of  the  first  order,  not  upon  the  value  of  the 
figures  at  the  right  of  it.  Thus,  in  425  and  400,  the  value 
of  the  4  is  the  same  in  both  numbers,  being  4  units  of  the 
third  order,  or  4  hundred. 

Care  should  be  taken  not  to  mistake  the  local  value  of  a  figure  for  the  value  of  the 
whole  number.  For,  although  the  value  of  the  4  (huudreflg)  is  the  same  in  the  two 
numbers.  425  and  400,  the  value  of  the  whole  of  the  first  number  is  greater  than  that 
of  the  second. 

5th.  Every  period  contains  three  figures  (units,  tens,  and 
hundreds),  except  the  left  hand  period,  which  sometimes 
contains  only  one  or  two  figures  (units,  or  units  and  tens). 

33.  As  all  the  principles  upon  which  the  writing  and 
reading  of  whole  numbers  depend  have  now  been  analyzed, 
we  will  present  these  principles  in  the  form  of  rules. 

Rfle  for  Notation. 

I.  Beginning  at  the  left  hand,  write  tlie  figures  belonging 
to  the  highest  2^eriod. 

II.  Write  the  hundreds,  tens,  and  units  of  each  successive 
period  in  their  order,  placing  a  cipher  wherever  an  order 
of  units  is  omitted. 

Rule  for  Numeration. 

I.  Separate  the  number  into  periods  of  three  figures  each, 
commencing  at  the  right  hand. 

II.  Beginning  at  the  left  hand,  read  each  period  separately, 
and  give  the  name  to  each  period,  except  the  last,  or  period 
of  units. 

34,  Until  the  pupil  can  write  numbers  readily,  it  may  be 
well  for  him  to  write  several  periods  of  ciphers,  point  them 
off,  and  over  each  period  write  its  name  ;  thus. 

Trillions,     Billions,      Millions,    Thousands,     Units. 

000,     000,     000,     000,     000, 

Fourth  principle  ?  What  caution  is  given  ?  Fifth  principle  ?  Rule 
for  notation  ?    Numeration  ? 


KOTATIOK     AKD     K  U  M  E  R  ATI  0  K .  19 

and   then  write  the  given  numbers  underneath,  in  their 
appropriate  places. 

Exercises  i:n"  Notation  akd  Numeratioi^-. 

Express  the  following  numbers  by  figures: — 

1.  Four  hundred  thirty-six. 

2.  Seven  thousand  one  hundred  sixty-four. 

3.  Twenty-six  thousand  twenty-six. 

4.  Fourteen  thousand  two  hundred  eighty. 

5.  One  hundred  seventy-six  thousand. 

6.  Four  hundred  fifty  thousand  thirty-nine. 

7.  Mnety-five  million. 

8.  Four  hundred  thirty-three  million  eight  hundred  six- 
teen thousand  one  hundred  forty-nine. 

9.  Mne  hundred  thousand  ninety. 

10.  Ten  million  ten  thousand  ten  hundred  ten. 

11.  Sixty-one  billion  five  million. 

12.  Five  trillion  eighty  billion  nine  million  one. 

Point  off,  numerate,  and  read  the  following  numbers: — 


13. 

8240. 

17. 

1010. 

21. 

370005. 

14. 

400900. 

18. 

57468139. 

22. 

9400706342. 

15. 

308. 

19. 

5628. 

23. 

38429526. 

16. 

60720. 

20. 

850026800. 

24. 

74268113759. 

25.  Write  seven  million  thirty-six. 

26.  Write  five  hundred  sixty-three  thousand  four. 

27.  Write  one  million  ninety-six  thousand. 

28.  Numerate  and  read  9004082501. 

29.  Numerate  and  read  2584503962047. 

30.  A  certain  number  contains  3  units  of  the  seventh 
order,  6  of  the  fifth,  4  of  the  fourth,  1  of  the  third,  5  of 
the  second,  and  2  of  the  first  ;  what  is  the  number  ? 

31.  What  orders  of  units  are  contained  in  the  number 
290648  ? 

32.  What  orders  of  units  are  contained  in  the  numbei 
1037050  ? 


20  SIMPLE     NUMBEES. 

ADDITION 
Mektal  Exercises. 

35.  1.  Henry  gave  5  dollars  for  a  vest,  and  7  dollars 
for  a  coat ;  what  did  lie  pay  for  both  ? 

Analysis.  He  gave  as  many  dollars  as  5  dollars  and  7  doUarsr 
which  are  12  dollars. 

2.  A  farmer  sold  a  pig  for  3  dollars,  and  a  calf  for  8  dol- 
lars ;  what  did  he  receive  for  both  ? 

3.  A  drover  bought  5  sheep  of  one  man,  9  of  another, 
and  3  of  another  ;  how  many  did  he  buy  in  all  ? 

4.  How  many  are  2  and  6  ?  2  and  7  ?  2  and  9  ?  2  and  8  ? 
2  and  10  ? 

5.  How  many  are  4  and  5  ?  4  and  8  ?  4  and  7  ?  4  and  9  ? 

6.  How  many  are  6  and  4?  6  and  6?  6  and  9  ?  6  and  7? 

7.  How  many  are  7  and  7  ?  7and0?  7and8?  7andlO? 
7  and  9  ? 

8.  How  many  are  5  and  4  and  6  ?  7  and  3  and  8  ?  6  and 
9  and  5  ? 

36.  From  the  preceding  operations  we  learn  that 
Addition  is  the  process  of  uniting  several  numbers  of 

the  same  kind  into  one  equivalent  number. 

37.  The  Sum  or  Amount  is  the  result  obtained  by  the 
process  of  addition. 

38.  The  sign,  +,  is  called  plus,  which  signifies  more. 
When  placed  between  two  numbers,  it  denotes  that  they  are 
to  be  added  ;  thus,  6  -(-  4,  shows  that  6  and  4  are  to  be  added. 

39.  The  sign,  =,  is  called  the  sign  of  equality.  When 
placed  between  two  numbers,  or  sets  of  numbers,  it  signifies 
that  they  are  equal  to  each  other  ;  thus,  the  expression 
64-4=10,  is  read  6  plus  4  is  equal  to  10,  and  denotes  that 
the  numbers  6  and  4,  taken  together,  equal  the  number  10. 

Define  addition.  The  sum  or  amount?  Sign  of  addition?  Of 
equality  ? 


ADDITIOiq". 


21 


Case  I. 

40.  "When  the  amount  of  each  colmnn  is  less 
than  10. 

1.  A  farmer  sold  some  hay  for  102  dollars,  six  cows  for 
162  dollars,  and  a  horse  for  125  dollars;  what  did  he 
receive  for  all  ? 

OPERATION.  Analysis.    Arrange  the  numbers  so  that 

4r^m  units  of  like  order  shall  stand  in  the  same 

column.  Then  add  the  columns  separately, 
for  convenience  commencing  at  the  right  hand, 
and  write  each  result  under  the  column  added. 
Thus,  we  have  5  and  2  and  2  are  9,  the  sum  of 
the  units ;  2  and  6  are  8,  the  sum  of  the  tens ; 
1  and  1  and  1  are  3,  the  sum  of  the  hundreds. 
Hence,  the  entire  amount  is  3  hundreds  8  tens 
and  9  units,  or  389. 

Examples  for  Pkactice. 


Amonnt, 


102 
162 
125 

389 


(3.) 

(3.) 

(4.) 

(5.) 

Pounds. 

Rods. 

Cents. 

Days. 

132 

245 

312 

437 

243 

321 

243 

140 

324 

132 

412 

321 

Ans.  699 

6.  What  is  the  sum  of  144,  321,  and  232  ?         Ans,  697. 

7.  What  is  the  amount  of  122,  333,  and  401  ?  Ans.  856. 

8.  What  is  the  sum  of  42,  103,  321,  and  32  ?    Ans.  498. 

9.  A  drover  bought  three  droves  of  sheep.  The  first  con- 
tained 230,  the  second  425,  and  the  third  340 ;  how  man;y 
sheep  did  he  buy  in  all  ?  Ans.  995. 

Case  H. 

41.  "When  the  amount  of  any  column  equals  or 
exceeds  10. 

1.  A  merchant  pays  725  dollars  a  year  for  the  rent  of  a 

Case  I  is  what  ?    Give  explanation.    Case  II  is  what  ? 


OPERATION. 

725 

475 

367 

Sum  of  the  units. 

17 

Sum  oi  ihe  tens, 

15 

Sum  of  the  hundreds, 

14 

Total  amount, 

1567 

22  SIMPLENUMBEKS 

store,  475  dollars  for  a  clerk,  and  367  dollars  for  other 
expenses  ;  what  is  the  amount  of  his  expenses  ? 

Analysis.  Arranging  the  num- 
bers as  in  Case  I,  we  first  add  the 
column  of  units,  and  find  the  sum  to 
be  17  units,  which  is  1  ten  and  7 
units.  Write  the  7  units  in  the  place 
of  units,  and  the  1  ten  in  the  place  of 
tens.  The  sum  of  the  figures  in  the 
column  of  tens  is  15  tens,  which  is  1 
hundred,  and  5  tens.  Write  the 
5  tens  in  the  place  of  tens,  and  the 
1  hundred  in  the  place  of  hundreds. 
Next  add  the  column  of  hundreds, 
and  find  the  sum  to  be  14  hundreds, 
which  is  1  thousand  and  4  hundreds.  Write  the  4  hundreds  in  the 
place  of  hundreds,  and  1  thousand  in  the  place  of  thousands.  Lastly, 
by  uniting  the  sum  of  the  units  with  the  sums  of  the  tens  and  hun- 
dreds, we  find  the  total  amount  to  be  1  thousand  5  hundreds  6  tens 
7  units,  or  1567. 

This  example  may  be  performed  by  another  methoa, 
which  is  the  common  one  in  practice.     Thus : 

OPERATION.        Analysis.    Arranging  the  numbers  as  before,  we  add 
725  the  first  column  and  find  the  sum  to  be  17  units  ;  writ- 

A>VK  ing  the  7  units  under  the  column  of  units,  we  add  the 

o/,«  1  ten  to  the  column  of  tens,  and  find  the  sum  to  be  16 

L         tens ;  writing  the  6  tens  under  the  column  of  tens,  we 

1567  add  the  1  hundred  to  the  column  of  hundreds,  and  find 

the  sum  to  bo  15  hundreds  ;  as  this  is  the  last  column, 
write  down  its  amount,  15 ;  and  the  whole  amount  is  1567,  as  before. 

1.  Units  of  the  same  order  are  written  in  the  same  column ;  and  when  the  sum 
in  any  column  is  10  or  more  than  10,  it  produces  one  or  more  units  of  a  hiprher  order, 
which  must  be  added  to  the  next  column.  This  process  is  sometimes  called  "  car- 
rying the  tens." 

2.  In  adding,  learn  to  pronounce  the  partial  results  without  naming  the  numbers 
sepai-ately ;  thus,  instead  of  saying  7  and  5  are  12,  and  5  are  17,  simply  pronounce 
the  results,  7,  12, 17,  etc. 

Give  explanation.  Second  explanation.  What  is  meant  by  carry- 
ing the  tens  ? 


ADDITIOl^.  23 

42.  From  the  preceding  examples  and  illustrations  we 
deduce  the  following 

EuLE.  I.  Write  the  numbers  to  he  added  so  that  the  units 
of  the  same  order  shall  stand  in  the  same  column  j  that  is, 
units  under  units,  tens  under  tens,  etc. 

II.  Commencing  at  units,  add  each  column  separately,  and 
write  the  sum  underneath,  if  it  he  less  than  ten. 

IIL  If  the  sum  of  any  column  he  ten  or  more  than  ten,  write 
the  unit  figure  only,  and  add  the  ten  or  tens  to  the  next  column, 

lY.  Write  the  entire  sum  of  the  last  colmnn, 

Pboof.  1st.  Begin  with  the  right  hand  or  unit  column, 
and  add  the  figures  in  each  column  in  an  opposite  direction 
from  that  in  which  they  were  first  added  ;  if  the  two  results 
agree,  the  work  is  supposed  to  be  right.     Or, 

2d.  Separate  the  numbers  added  into  two  sets,  by  a  hori- 
zontal line ;  find  the  sum  of  each  set  separately  ;  add  these 
sums,  and  if  the  amount  be  the  same  as  that  first  obtained, 
the  work  is  presumed  to  be  correct. 

By  the  methods  of  proof  here  given,  the  numbers  are  united  in  new  combina- 
tione,  which  render  it  ahnost  impossible  for  two  precisely  similar  mistakes  to 
occur. 

The  first  method  is  the  one  commonly  used  in  business. 


Examples  for  Practice. 


(2.) 

(3.) 

(4.) 

(5.) 

(6.) 

lOles. 

Inches. 

Tons. 

Feet. 

Bushels. 

24 

321 

427 

1342 

3420 

48 

479 

321 

7306 

7021 

96 

165 

903 

6254 

327 

82 

327 

278 

8629 

97 

250 

1292 

1929 

22531 

10865 

Rule,  first  step ?    Second?    Third?    Fourth?    Proof, first  method  1 
Second  ?    Upon  what  principle  are  these  methods  of  proof  founded  ? 


u 


SIMPLE 

I^UMBERS. 

(7.) 

(8.) 

(9.) 

(10.) 

lours. 

Years. 

Gallons. 

Rods. 

347 

7104 

3462 

47637 

506 

3762 

863 

3418 

218 

9325 

479 

703 

312 

4316 

84 

26471 

4:24. 

2739 

57 

84 

11.  42  +  64+98  +  70  +  37  =  how  many?        Ans.  311. 

12.  312  +  425  + 107  +  391  +  76  =  how  many  ? 

Ans,  1311. 

13.  1476  +  375  +  891  +  66  +  80  =  how  many? 

Ans.  2888. 

14.  37042  +  1379  +  809  +  127  +  40  =  how  many? 

^/^5.  39397. 

15.  "What  is  the  sum  of  one  thousand  six  hundred  fifty- 
six,  eight  hundred  nine,  three  hundred  ten,  and  ninety-four? 

Ans.  2869. 

16.  Add  forty-two  thousand  two  hundred  twenty,  ten 
thousand  one  hundred  five,  four  thousand  seventy-five,  and 
five  hundred  seven.  Ans,  56907. 

17.  Add  two  hundred  ten  thousand  four  hundred,  one 
hundred  thousand  five  hundred  ten,  ninety  thousand  six 
hundred  eleven,  forty-two  hundred  twenty-five,  and  eight 
hundred  ten.  Ans.  406556. 

18.  What  is  the  sum  of  the  following  numbers :  seventy- 
five,  one  thousand  ninety-five,  six  thousand  four  hundred 
thirty-five,  two  hundred  sixty-seven  thousand,  one  thousand 
tour  hundred  fifty-five,  twenty-seven  million  eighteen,  two 
hundred  seventy  million  twenty-seven  thousand  ? 

Ans.  297303078. 

19.  A  man  on  a  journey  traveled  the  first  day  37  miles, 
the  second  33  miles,  the  third  40  miles,  and  the  fourth 
35  miles ;  how  far  did  he  travel  in  the  four  days  ? 

20.  A  wine  merchant  has  in  one  cask  75  gallons,  in 
another  65,  in  a  third  57,  in  a  fourth  83,  in  a  fifth  74,  and 
in  a  sixth  67  ;  how  many  gallons  has  he  in  all  ?    Ans-  421 


ADDITION.  25 

21.  An  estate  is  to  be  shared  equally  by  four  heirs,  and 
the  portion  to  each  heir  is  to  be  3754  dollars  ;  what  is  the 
amount  of  the  estate  ?  Ans.  15016  dollars. 

22.  How  many  men  in  an  army  consisting  of  52714  in- 
fantry, 5110  cavalry,  6250  dragoons,  3927  light-horse,  928 
artillery,  250  sappers,  and  406  miners  ? 

23.  A  merchant  deposited  56  dollars  in  a  bank  on  Mon- 
day, 74  on  Tuesday,  120  on  Wednesday,  96  on  Thursday, 
170  on  Friday,  and  50  on  Saturday;  how  much  did  he 
deposit  during  the  week  ? 

24.  A  merchant  bought  at  public  sale  746  yards  of  broad- 
cloth, 650  yards  of  muslin,  2100  yards  of  flannel,  and  250 
yards  of  silk  ;  how  many  yards  in  all  ? 

25.  Eive  persons  deposited  money  in  the  same  bank ;  the 
first,  5897  dollars;  the  second,  12980  dollars;  the  third, 
65973  dollars;  the  fourth,  37345  dollars;  and  the  fifth  as 
much  as  the  first  and  second  together ;  how  many  dollars 
did  they  aU  deposit  ?  Ans,  141072  dollars. 

26.  A  man  willed  his  estate  to  his  wife,  two  sons,  and 
four  daughters ;  to  his  daughters  he  gave  2630  dollars  apiece, 
to  his  sons,  each  4647  dollars,  and  to  his  wife  3595  dollars ; 
how  much  was  his  estate  ?  Ans.  23409  dollars. 

(27.)  (28.)  (29.)  (30.)  (31.) 


476 

908 

126 

443 

180 

390 

371 

324 

298 

976 

915 

569 

503 

876 

209 

207 

245 

891 

569 

314 

841 

703 

736 

137 

563 

632 

421 

517 

910 

842 

234 

127 

143 

347 

175 

143 

354 

274 

256 

224 

536 

781 

531 

324 

135 

245 

436 

275 

463 

253 

R.P. 


26  SIMPLE     NUMBERS. 

32.  A  man  commenced  farming  at  the  west,  and  raised, 
the  first  year,  724  bushels  of  corn ;  the  second  year,  3498 
bushels;  the  third  year,  9872  bushels;  the  fourth  year, 
9964  bushels;  the  fifth  year,  11078  bushels;  how  many 
bushels  did  he  raise  in  the  five  years  ?    Ans.  35136  bushels. 

33.  A  has  3648  dollars,  B  has  7035  dollars,  C  has  429 
dollars  more  than  A  and  B  together,  and  D  has  as  many 
dollars  as  all  the  rest;  how  many  dollars  has  D?  How 
many  have  all  ?  Ans,  All  have  43590  dollars. 

34.  A  man  bought  three  houses  and  lots  for  15780  dollars, 
and  sold  them  so  as  to  gain  695  dollars  on  each  lot ;  for  how 
much  did  he  sell  them?  Ans.  17865  dollars. 

35.  At  the  battle  of  Waterloo,  which  took  place  June  18th, 
18 ^'5,  the  estimated  loss  of  the  French  was  40000  men;  of 
the  Prussians,  38000 ;  of  the  Belgians,  8000 ;  of  the  Hano- 
verians, 3500;  and  of  the  English,  12000;  what  was  the 
entire  loss  of  life  in  this  battle  ? 

36.  The  expenditures  for  educational  purposes  in  New 
England  for  the  year  1850  were  as  follows :  Maine,  380623 
dollars;  New  Hampshire,  221146  dollars ;  Vermont, 246604 
dollars;  Massachusetts,  1424873  dollars;  Rhode  Island, 
136729  dollars;  and  Connecticut,  430826  dollars;  what  was 
the  total  expenditure  ?  Ans.  2840801  dollars, 

37.  The  eastern  continent  contains  31000000  square 
miles;  the  western  continent,  13750000  ;  Australia,  Green- 
land, and  other  islands,  5250000 ;  what  is  the  entire  area  of 
the  land  surface  of  the  globe  ? 

38.  The  population  of  New  York,  in  1850,  was  515547 ; 
Boston,  136881;  Philadelphia,  340045;  Chicago,  29963; 
St.  Louis,  77860;  New  Orleans,  116375;  what  was  the 
entire  population  of  these  cities  ?  Ans,  1216671. 

39.  The  population  of  the  globe  is  estimated  as  follows: 
North  America,  39257819  ;  South  America,  18373188  ;  Eu- 
rope, 265368216;  Asia,  630671661;  Africa,  61688779; 
Oceanica,  23444082 ;  what  is  the  total  population  of  the 
globe  according  to  this  estimate  ?  Ans,  1038803745. 


ADDITION 


27 


40.  The  railroad  distance  from  New  York  to  Albany  is 
144  miles ;  from  Albany  to  Buffalo,  298 ;  from  Buffalo  to 
Cleveland,  183  ;  from  Cleveland  to  Toledo,  109  ;  from  To- 
ledo to  Springfield,  365 ;  and  from  Springfield  to  St.  Louis, 
95  miles ;  what  is  the  distance  from  New  York  to  St.  Louis  ? 

41.  A  man  owns  farms  valued  at  56800  dollars ;  city  lots 
Valued  at  86760  dollars ;  a  house  worth  12500  dollars,  and 
other  property  to  the  amount  of  6785  dollars  ;  what  is  the 
entire  value  of  his  property  ?  Ans.  162845  dollars. 


(42.) 

(43.) 

(44.) 

(45.) 

15038 

26881 

41919 

93808 

7404 

12173 

19577 

41371 

34971 

39665 

74736 

110525 

30359 

33249 

66768 

102936 

6293 

6318 

12673 

17087 

2875 

4318 

7193 

13251 

16660 

34705 

61365. 

112110 

64934 

80597 

155497 

220619 

80901 

95299 

183134 

225255 

7444 

8624 

16845 

68940 

57068 

53806 

111139 

176974 

17255 

18647 

.   35902 

86590 

32543 

41609 

82182 

149162 

40022 

•  35077 

75153 

109355 

56063 

46880 

132936 

283910 

33860 

41842 

82939 

112511 

17548 

26876 

44424 

72908 

28944 

36642 

65586 

157672 

16147 

29997 

52839 

86160 

38556 

44305 

83211 

119557 

234882 

262083 

522294 

839398 

39058 

39744 

78861 

117787 

152526 

169220 

353428 

471842 

179122 

198568 

386214 

671778 

7626 

8735 

17005 

41735 

1218099 

1395860 

28  SIMPLE     NUMBEK8. 

SUBTEAOTIOTsT. 

Mental  Exeecises. 

43.  1.  A  farmer,  having  14  cows,  sold  6  of  them ;  how 
many  had  he  left  ? 

ANAiiYSis.    He  had  as  many  left  as  14  cows  less  6  cows,  which  are 
8  cows. 

2.  Stephen,  haying  9  marbles,  lost  4  of  them ;  how  many 
had  he  left  ? 

3.  If  a  man  earn  10  dollars  a  week,  and  spend  6  dollars 
for  provisions,  how  many  dollars  has  he  left  ? 

4.  A  merchant,  having  16  barrels  of  flour,  seUs  9  of  them ; 
how  many  barrels  has  he  left  ? 

5.  Charles  had  18  cents,  and  gave  10  of  them  for  a  book; 
how  many  cents  had  he  left  ? 

6.  James  is  17  years  old,  and  his  sister  Julia  is  5  years 
younger ;  how  old  is  Julia  ? 

7.  A  grocer,  having  20  boxes  of  lemons,  sold  11  boxes; 
how  many  boxes  had  he  left  ? 

8.  From  a  cistern  containing  25  barrels  of  water,  15  bar- 
rels leaked  out ;  how  many  barrels  remained  ? 

9.  Paid  16  dollars  for  a  coat,  and  7  dollars  for  a  vest ; 
how  much  more  did  the  coat  cost  than  the  vest  ? 

10.  How  many  are  18  less  5  ?  17  less  8  ?    12  less  7  ? 

11.  How  many  are  20  less  14  ?  18  less  12  ?     19  less  11  ? 

12.  How  many  are  11  less  3  ?  16  less  11  ?     19  less  8  ? 
20  less  9  ?    22  less  20  ? 

44.  Subtraction  is  the  process  of  finding  the  difference 
between  two  numbers  of  the  same  kind. 

45.  The  Minuend  is  the  number  to  be  subtracted  from. 

46.  The  Subtrahend  is  the  number  to  be  subtracted. 

Define  subtraction  ?    Minuend  ?    Subtrahend  t 


SUBTEACTIOK.  29 

47.  The  Diflference  or  Kemainder  is  the  result  ob- 
tained by  the  process  of  subtraction. 

The  minuend  and  subtrahend  must  be  like  numbers ;  thus,  5  dollars  from  9  dol- 
lars leave  4  dollars ;  5  apples  from  9  apples,  leave  4  apples ;  but  it  would  be  absurd 
to  say  5  apples  from  9  dollars,  or  5  dollars  from  9  apples. 

48.  The  sign,  — ,  is  called  minus ,  which  signifies  less. 
When  placed  between  two  numbers,  it  denotes  that  the  one 
after  it  is  to  be  taken  from  the  one  before  it.  Thus,  8—6 
=2  is  read  8  minus  6  equals  2,  and  shows  that  6,  the  sub- 
trahend, taken  from  8,  the  minuend,  equals  2,  the 
remainder. 

Case  I. 

49.  "When  no  figure  in  the  subtrahend  is  greater 
than  the  corresponding  figure  in  the  minuend. 

1.  From  574  take  323. 

Analysis.  Write  the  less  number  un- 
der the  greater,  with  units  under  units, 
tens  under  tens,  etc.,  and  draw  a  line 
underneath.  Then,  beginning  at  the  right 
hand,  subtract  separately  each  figure  of 
the  subtrahend  from  the  figure  above  it  in 
the  minuend.  Thus,  3  from  4  leaves.  1,  which  is  the  difference  of  the 
units ;  2  from  7  leaves  5,  the  difference  of  the  tens  ;  3  from  5  leaves  2, 
the  difference  of  the  hundreds.  Hence,  we  have  for  the  whole  differ- 
ence, 2  hundreds  5  tens  and  1  unit,  or  251. 


Examples  foe  Peactice. 


OPERATION. 

Minuend, 

574 

Subtrahend, 

323 

Remainder, 

251 

Minuend, 

(2.) 

876 

(3.) 

676 

(4.) 
367 

(5.) 
925 

Subtraliend, 
Remainder, 

334 

542 

415 
261 

152 
215 

213 

712 

Case  I  is  wliat  ?     Give  explanation. 


30  SIMPLENUMBERS 


(6.)        (7.) 
From    876        732 

(8.) 

987 

(9.) 
498 

Take     523        522 

782 

178 

10.  From  3276  take  2143. 

Remainders. 
1133. 

11.  From  7634  take  3132. 

4502. 

12.  From  41763  take  11521. 

30242. 

13.  From  18346  take  5215. 

13131. 

14  From  397631  take  175321. 

222310. 

15.  Subtract  47321  from  69524. 

22203. 

16.  Subtract  16330  from  48673. 

32343. 

17.  Subtract  291352  from  895752. 

604400. 

18.  Subtract  84321  from  397562. 

313241. 

19.  A  farmer  paid  645  dollars  for  a  span  of  horses  and 
a  carriage,  and  sold  them  for  522  dollars;  what  did  he 
lose? 

20.  A  man  bought  a  mill  for  3724  dollars,  and  sold  it  for 
4856  dollars;  what  did  he  gain?  Ans,  1132  dollars. 

21.  A  drover  bought  1566  sheep,  and  sold  435  of  them , 
how  many  had  he  left  ?  Ans.  1131  sheep. 

22.  A  piece  of  land  was  sold  for  2945  dollars,  which  was 
832  dollars  more  than  it  cost ;  what  did  it  cost  ? 

23.  A  gentleman  willed  to  his  son  15768  dollars,  and  to 
his  daughter  4537  dollars;  how  much  more  did  he  will  to 
the  son  than  to  the  daughter  ?  Ans.  11231  dollars. 

24.  A  merchant  sold  goods  to  the  amount  of  6742  dollars, 
and  by  so  doing  gained  2540  dollars ;  what  did  the  goods 
cost  him  ? 

25.  If  I  borrow  15475  dollars  of  a  person,  and  pay  him 
4050  dollars,  what  do  I  still  owe  him  ? 

26.  In  1850  the  white  population  of  the  United  States 
was  19,553,068,  and  the  slave  population  3,204,313;  what 
was  the  difference  ? 

27.  The  population  of  Great  Britain  in  1851  was  20,936,468, 
and  of  England  alone,  16,921,888 ;  what  was  the  difference  ? 


subtraction.  31 

Case  IL 

50.  "When  any  figure  in  the  subtrahend  is  greater 
than  the  corresponding  figure  in  the  minuend. 

1.  From  846  take  359. 

OPERATION.  Analysis.    In  tMs  example  we 

(7)    (13)    (16)  cannot  take  9  units  from  6  units. 

Minuend,           8       4       6  From  the  4  tens  we  take  1  ten,  which 

Subtrahend.       3       5        9  equals  10  units,  and  add  to  the  6 

units,  making  16  units ;  9  units  from 

Eemaiuder,        4       8       7  ^q  ^^^^^  ^^^^^  7  ^i^s,  which  we 

write  in  the  remainder  in  units'  place.  As  we  have  taken  1  ten  from 
the  4  tens,  3  tens  onlj  are  left.  We  cannot  take  5  tens  from  3  tens  ; 
so  from  the  8  hundreds  we  take  1  hundred,  which  equals  10  tens,  and 
add  to  the  3  tens,  making  13  tens ;  5  tens  from  13  tens  leave  8  tens, 
which  we  write  in  the  remainder  in  tens'  place.  As  we  have  taken 
1  hundred  from  the  8  hundreds,  7  hundreds  only  are  left ;  3  hundreds 
from  7  hundreds  leave  4  hundreds,  which  we  write  in  the  remainder 
in  hundreds'  place,  and  we  have  the  whole  remainder,  487. 

The  numbers  written  over  the  minuend  are  used  simply  to  explain  more  clearly 
the  method  of  subtracting ;  in  practice  the  process  should  be  performed  mentally, 
and  these  numbers  omitted. 

The  following  method  is  more  in  accordance  with  prac- 
tice. 

OPERATION.  Analysis.    Since  we  cannot  take  9  units  from  6 

®  ^  ^  units,  we  add  10  units  to  6  units,  making  16  units ; 

§  §1  9  units  from  16  units  leave  7  units.     But  as  we  have 

QAf*  added  10  units,  or  1  ten,  to  the  minuend,  we  shall 

have  a  remainder  1  ten  too  large,  to  avoid  which,  we 
add  1  ten  to  the  5  tens  in  the  subtrahend,  making  6 
48  7  *®^s.     We  can  not  take  6  tens  from  4  tens ;  so  we 

add  10  tens  to  4,  making  14  tens;  6  tens  from  14 
tens  leave  8  tens.  Now,  having  added  10  tens,  or  1  hundred,  to  the 
minuend,  we  shall  have  a  remainder  1  hundred  too  large,  unless  we 
add  1  hundred  to  the  3  hundreds  in  the  subtrahend,  making  4  hun- 
dreds ;  4  hundreds  from  8  hundreds  leave  4  hundreds,  and  we  have 
for  the  total  remainder,  487,  the  same  as  before. 

Case  n  is  what  ?    Give  explanation.    Second  explanation. 


32  SIMPLE    NUMBERS. 

The  process  of  adding  10  to  the  minuend  is  sometimes  called  borrowing  10,  and 
that  of  adding  1  to  the  next  figure  of  the  subtrahend,  carrying  one, 

•   51.  From  the  preceding  illustrations  we  have  the  fol- 
lowing 

Rule.  I.  Write  the  less  number  under  the  greater,  plac- 
ing units  of  the  same  order  in  the  same  column, 

II.  Begifi  at  the  right  hand,  and  take  each  figure  of  the 
subtrahend  from  the  figure  above  it,  and  write  the  result  un- 
derneath. 

III.  If  any  figure  in  the  subtrahend  be  greater  than  the 
corresponding  figure  above  it,  add  10  to  that  upper  figure  be- 
fore subtracting,  and  then  add  1  to  the  next  left  hand  figure 
of  the  subtrahend. 

Pkoof.  Add  the  remainder  to  the  subtrahend,  and  if 
their  sum  be  equal  to  the  minuend,  the  work  is  supposed 
to  be  right. 

Examples  fob  Peactice. 


Minuend, 

(2.) 

873 

(3.) 
7432 

(4.) 
1969 

(5.) 
8146 

Subtrahend, 

538 

6711 

1408 

4377 

Eemainder, 

335 

From 

(6.) 

Gallons. 

3176 

(7.) 

Bushels. 

9076 

(8.) 

MUes. 

7320 

(9.) 

Days. 

6097 

Take 

2907 

4567 

3871 

3809 

From 

(10.) 

Dollars. 

76377 

(11.) 

Rods. 

67777 

(12.) 

Acres. 

900076 

(13.) 

Feet 

767340 

Take 

45761 

46699 

899934 

5039 

What  do  we  mean  by  borrowing  10  ?    By  carrying  ?    Rule,  first  step  1 
Second?    Third?    Proof? 


SUBTRACTION 


33 


14.  479—382  =  how  many  ?  Ans.  97. 

15.  6593—1807  =  how  many?  Ans.  4786. 

16.  17380—3417  =  how  many?  Ans.  13963. 

17.  80014— 43190  =  how  many?  Ans.  36824. 

18.  282731—90756  =  how  many?  Ans.  191975. 

19.  From  234100  take  9970. 

20.  From  345673  take  124799. 

21.  From  4367676  take  256569.  Ans.  4111107 

22.  From  3467310  take  987631.  Ans.  2479679. 

23.  From  941000  take  5007.  Afis.  935C93. 

24.  From  1970000  take  1361111.  A7is.  608889. 

25.  From  290017  take  108045. 

26.  Take  3077097  from  7045676.  Ans.  3968579. 

27.  Take  9999999  from  60000000.  Ans.  50000001. 

28.  Take  220202  from  4040053.  Ans,  3819851. 

29.  Take  2199077  from  3000001.  Ans.  800924. 

30.  Take  377776  from  8000800.  Ans.  7623024. 

31.  Take  501300347  from  1030810040. 

32.  Subtract  nineteen  thousand  nineteen  from  twenty 
thousand  ten.  Ans.  991. 

33.  From  one  million  nine  thousand  six  take  twenty  thou- 
sand four  hundred.  Ans.  988606. 

34.  What  is  the  difference  between  two  million  seven 
thousand  eighteen,  and  one  hundred  five  thousand  seven- 
teen? 

Examples  Combiking  Addition  and  Subtraction. 

52,  1.  A  merchant  gave  his  note  for  5200  dollars.  He 
paid  at  one  time  2500  dollars,  and  at  another  175  dollars  ; 
what  remained  due  ?  Ans.  2525  dollars. 

2.  A  traveler  who  was  1300  miles  from  home,  traveled 
homeward  235  miles  in  one  week,  in  the  next  275  miles,  in 

.the  next  325  miles,  and  in  the  next  280  miles ;  how  far  had 
he  still  to  go  before  he  would  reach  home  ?    Ans.  185  miles. 

3.  A  man  deposited  in  bank  8752  dollars  ;  hs  drew  out  at 
one  time  4234  dollars,  at  another  1700  dollars,  at  another  962 


34  SIMPLE     NUMBERS. 

dollars,  and  at  another  49  dollars ;  how  much  had  he  remain- 
ing in  bank?  Ans.  1807  dollars. 

4.  A  man  bought  a  farm  for  4765  dollars,  and  paid  750 
dollars  for  fencing  and  other  improvements  ;  he  then  sold 
it  for  384  dollars  less  than  it  cost  him ;  what  did  he  receive 
for  it  ?  Ans,  5131  dollars. 

5.  A  forwarding  merchant  had  in  his  warehouse  7520  bar 
rels  of  flour  ;  he  shipped  at  one  time  1234  barrels,  at  anothei 
time  1500  barrels,  and  at  another  time  1805  barrels ;  how 
many  barrels  remained  ? 

6.  A  had  450  sheep,  B  had  175  more  than  A,  and  C 
had  as  many  as  A  and  B  together  minus  114  ;  how  many 
sheep  had  C  ?  Ans.  961  sheep. 

7.  A  farmer  raised  1575  bushels  of  wheat,  and  900  bushels 
of  com.  He  sold  807  bushels  of  wheat,  and  391  bushels  of 
com  to  A,  and  the  remainder  to  B  ;  how  much  of  each  did 
he  sell  to  B  ?  Ans.  768  bushels  of  wheat,  and  509  of  com. 

8.  A  man  traveled  6784  miles ;  2324  miles  by  railroad, 
1570  miles  in  a  stage  coach,  450  miles  on  horseback,  175 
miles  on  foot,  and  the  remainder  by  steamboat ;  how  many 
miles  did  he  travel  by  steamboat  ?  Ans.  2265  miles. 

9.  Three  persons  bought  a  hotel  valued  at  35680  dollars. 
The  first  agreed  to  pay  7375  dollars,  the  second  agreed  to 
pay  twice  as  much,  and  the  third  the  remainder  ;  Avhat  was 
the  third  to  pay  ?  Ans.  13555  dollars. 

10.  Borrowed  of  my  neighbor  at  one  time  750  dollars,  at 
another  time  379  dollars,  and  at  another  450  dollars.  Having 
paid  him  1000  dollars,  how  much  do  I  still  owe  him  ? 

Ans.  579  dollars. 

11.  A  man  worth  6709  dollars  received  a  legacy  of  3000 
dollars,  ^e  spent  4379  dollars  in  traveling  ;  how  much  had 
he  left? 

12.  In  1850  the  number  of  white  males  in  the  United 
States  was  10026402,  and  of  white  females  9526666  ;  of 
these,  8786968  males,  and  8525565  females  were  native 
bom  ;  how  many  of  both  were  foreign  bom  ?  Ans.  2240535. 


MULTIPLICATION.  35 

M  D  LTIPLIC  ATIOI^. 

Mental  Exercises. 

53.  1.  What  will  4  pounds  of  sugar  cost,  at  8  cents  a 
pound  ? 

Analysis.  Four  pounds  will  cost  as  mucli  as  the  price,  8  cents 
taken  4  times ;  thus,  8  +  8  +  8  +  8=33.  But  instead  of  adding,  we  may 
say,— since  one  pound  cost  8  cents,  4  pounds  will  cost  4  times  8  cents, 
or  33  cents. 

2.  If  a  ream  of  paper  cost  3  dollars,  what  will  2  reams 
cost? 

3.  At  7  cents  a  quart,  what  will  4  quarts  of  cherries 
cost? 

4.  At  12  dollars  a  ton,  what  will  3  tons  of  hay  cost?  4 
tons  ?  5  tons  ? 

5.  There  are  7  days  in  1  week ;  how  many  days  in  6  weeks? 
In  8  weeks  ? 

6.  What  will  9  chairs  cost,  at  10  shillings  apiece  ? 

7.  If  Henry  earn  12  dollars  in  1  month,  how  much  can 
he  earn  in  5  months  ?  in  7  months  ?  in  9  months  ? 

8.  What  will  11  dozen  of  eggs  cost,  at  9  cents  a  dozen  ? 
At  10  cents  ?    At  12  cents  ? 

9.  When  flour  is  7  dollars  a  barrel,  what  must  be  paid  for 
7  barrels  ?  for  9  barrels  ?  for  12  barrels  ? 

10.  At  9  dollars  a  week,  what  will  4  weeks'  board  cost  ? 
7  weeks'  ?  9  weeks'  ? 

11.  If  I  deposit  12  dollars  in  a  sa\dngs  bank  every  month, 
how  many  dollars  will  I  deposit  in  6  months  ?  In  8  months  ? 
In  9  months  ? 

12.  At  9  cents  a  foot,  what  will  4  feet  of  lead  pipe  cost  ? 
.feet?    10  feet? 

13.  When  hay  is  8  dollars  a  ton,  how  much  will  3  tons 
cost  ?    4  tons  ?     7  tons  ?    9  tons  ?     11  tons  ? 


36 


SIMPLE     KUMBERS 


14  What  will  be  the  cost  of  11  barrels  of  apples,  at  2 
dollars  a  barrel  ?  at  3  doUarsi? 

15.  At  10  cents  a  pound,  what  will  9  pounds  of  sugar 
cost?  11  pounds?   12  pounds? 

54:,  Multiplication  is  the  process  of  taking  one  of  two 
given  numbers  as  many  times  as  there  are  units  in  the  other. 

55»  The  Multiplicand  is  the  number  to  be  multiplied. 

56.  The  Multiplier  is  the  number  by  which  to  multiply, 
and  shows  how  many  times  the  multiplicand  is  to  be  taken. 

57.  The  Product  is  the  result  obtained  by  the  process  of 
multiplication. 

5S,  The  Factors  are  the  multiplicand  and  multiplier. 

1.  Factors  are  producers,  and  the  multiplicand  and  multiplier  are  called  factors 
because  they  produce  the  product. 

2.  Multiplication  is  a  short  method  of  performing  addition  when  the  numbers  to 
be  added  are  equal. 

59.  The  sign,  x ,  placed  between  two  numbers,  denotes 
that  they  are  to  be  multiplied  together  ;  thus,  9  x  6=54,  is 
read  9  times  6  equals  54. 

Multiplication^  Table. 


Ix  1=  1 

2x    1=  2 

3x   1=  3 

4x   1=  4 

Ix   2=  2 

2x   2=  4 

3x    2=  6 

4x   2=  8 

Ix   3=  3 

2x   3=  6 

3x   3=  9 

4x   3  =  12 

Ix   4=  4 

2x   4=  8 

3x   4=12 

4x   4=16 

Ix   5=  5 

2x   5=10 

3x   5  =  15 

4x   5=20 

Ix    6=  6 

2x    6  =  12 

3x    6  =  18 

4x    6=24 

Ix   7=  7 

2x    7  =  14 

3x   7=21 

4x    7=28 

Ix   8=  8 

2x   8=16 

3x   8=24 

4x   8=3:i 

Ix    9=  9 

2x   9  =  18 

3x   9=27 

4x    9=36 

1x10=10 

2x10=20 

3x10=30 

4x10=40 

1x11=11 

2x11=22 

3x11=33 

4x11=44 

1x12  =  12 

2x12=24 

3x12=36 

4x12=48 

Define  multiplication.  Multiplicand.  Multiplier.  Product.  Fac- 
tor. Multiplication  la  a  abort  method  of  what  ?  What  is  the  sign  of 
multiplication  ? 


MULTIPLICATION. 


37 


5x   1=  5 

6x    1=  6 

7x    1=  7 

8x   1=  8 

5x   2  =  10 

6x    2  =  12 

7x   2=14 

8x   2=16 

5x   3zizl5 

6x   3  =  18 

7x   3=21 

8x   3=24 

5x   4=20 

6x   4=24 

7x   4=28 

8x   4=32 

5x   5=25 

6x   5=30 

7x   5=35 

8x   5=40 

5x    6=30 

6x    6=36 

7x    6=42 

8x    6=48 

5x    7=35 

6x    7=42 

7x    7=49 

8x    7  =  56 

ox    8=40 

6x   8=48 

7x   8=56 

8x    8=64 

5x    9=45 

6x    9-54 

7x    9  =  63 

8x    9  =  72 

5x10  =  50 

6x10=60 

7x10=70 

8x10=80 

5x11  =  55 

6x11  =  66 

7x11  =  77 

8x11=88 

5x12  =  60 

6x12  =  72 

7x12=84 

8x12=96 

9x 
9x 
9x 


1  = 
2= 
3  = 
4= 

5  = 

6  = 

7= 


X 
X 
X 
X 

X  8  = 
X  9  = 
xlO= 
xll  = 


9x12: 


9 
18 
27 
36 
45 
54 
63 
72 
81 
90 
99 
108 


10  X 
10  X 
10  X 
10  X 
10  X 
10  X 
10  X 
10  X 
10  X 


1  = 

2  = 

3  = 
4= 

5  = 

6  = 

7= 


10 
20 
30 
40 
50 
60 
70 
80 
90 


9  = 
10x10=100 
10x11  =  110 
10x12  =  120 


11  X 
11  X 
llx 
11  X 
llx 
llx 
llx 
llx 
llx 


1  = 

2  = 

3  = 
4= 

5  = 

6  = 
7= 


11 

22 
33 
44 
55 
66 
77 
88 
99 


9  = 
11x10=110 
11x11  =  121 
11x12=132 


12  X 
12  X 
12  X 
12  X 
12  X 
12  X 
12  X 
12  X 
12  X 


1  = 
2= 

3  = 

4  = 

5:= 

6  = 

7  = 

8  = 
9= 


12 
24 
36 

48 
60 
72 
84 
96 
108 


12x10  =  120 
12x11  =  132 
12x12=144 


OPERATION. 


Case  I. 
60.  When  the  multiplier  consists  of  one  figure. 
1.  Multiply  374  by  6. 

Analysis.  In  this  example  it  is 
required  to  take  374  six  times.  If  we 
take  the  units  of  each  order  6  times, 
we  shall  take  the  entire  number  6 
times.  Therefore,  writing  the  multi- 
plier under  the  unit  figure  of  the  mul- 
tiplicand, we  proceed  as  follows:  6 
times  4  units  are  24  units ;  6  times  7 
tens  are  42  tens ;  6  times  3  hundreds 
are  18  hundreds  ;  and  adding  these 
partial  products,  we  obtain  the  entire 
product,  2244. 


Multiplicand, 
Multiplier, 

Units, 
Tens, 
Hundreds, 
Product, 


374 
6 

24 

42 


2244 


Case  I  is  what  ?    Give  explanation. 


IMPLE    NUMBEES. 


The  operation  in  this  example  may  be  performed  in  an- 
other way,  which  is  the  one  in  common  use. 


OPEEATION. 

874 

6 

2244 


Akalysis.  Writing  the  numbers  as  before,  begin 
at  the  right  hand  or  unit  figure,  and  say :  6  times  4 
units  are  24  units,  which  is  2  tens  and  4  units;  write 
the  4  units  in  the  product  in  units'  place,  and  reserve 
the  2  tens  to  add  to  the  next  product ;  C  times  7  tent 
are  42  tens,  and  the  two  tens  reserved  in  the  last 
product  added,  are  44  tens,  which  is  4  hundreds  and  4  teTis  ;  write  the 
4  tens  in  the  product  in  tens'  place,  and  reserve  the  4  hundreds  to  add 
to  the  next  product  ;  6  times  3  hundreds  are  18  hundreds,  and  4  hun- 
dreds added  are  22  hundreds,  which  being  written  in  the  product  in 
the  places  of  hundreds  and  thousands,  gives,  for  the  entire  product, 
2244. 

61,  The  unit  value  of  a  number  is  not  changed  by  re- 
peating the  number.  As  the  multiplier  always  expresses 
times ^  the  product  must  have  the  same  unit  value  as  the 
multiplicand.  But  since  the  product  of  any  two  numbers 
will  be  the  same,  whichever  factor  is  taken  as  a  multiplier, 
either  factor  may  be  taken  for  the  multiplier  or  multiplicand. 

In  multiplying,  learn  to  pronounce  the  partial  results,  as  in  addition,  without 
naming  the  numbers  separately ;  thus,  in  the  last  example,  instead  of  saying  6  times 
4  are  24,  6  times  7  are  42  and  2  to  carry  are  44,  6  times  3  are  18  and  4  to  carry  are  22, 
pronounce  only  the  results,  24,  44,  22,  performing  the  operations  mentally.  This 
will  greatly  facilitate  the  process  of  multiplying. 


Examples  for  Practice. 


(2.) 

(3.) 

(4.) 

Multiplicand, 

7324 

6812 

34651 

Multipher, 

4 

6 

5 

Product, 

29296 

40872 

173255 

(5.) 

(6.) 

(7.)      • 

(8.) 

82456 

92714 

28093 

46247 

3 

7 

8 

9 

Second  explanation.     Repeating  a  number  has  what  effect  on  the 
unit  value  ?    The  product  must  be  of  the  same  kind  as  what  ? 


M  U  L  T  I  P  L  I  C  A  T  I  O  IsT . 


39 


9.  Multiply    32746  by  5. 

10.  Multiply  840371  by  7. 

11.  Multiply  137629  by  8. 

12.  Multiply    93762  by  3. 

13.  Multiply  543272  by  4. 

14.  Multiply  703164  by  9. 


Ans, 
Ans, 
Ans, 
Ans, 
Ans. 
Ans, 


163730. 

5882597. 
1101032. 
281286. 
2173088. 
6328476. 


15.  Wbat  will  be  the  cost  of  344  cords  of  wood  at  4  dol- 
lars a  cord  ?  Ans.  1376. 

16.  How  much  wiU  an  army  of  7856  men  receive  in  one 
week,  if  each  man  receive  6  dollars  ?    Ans,  47136  dollars. 

17.  In  one  day  are  86400  seconds  ;  how  many  seconds  in 
7  days  ?  Ans,  604800  seconds. 

18.  What  will  7640  bushels  of  wheat  cost,  at  9  shillings 
a  bushel  ?  Ans.  68760  shillings. 

19.  At  5  dollars  an  acre,  what  will  2487  acres  of  land 
cost  ?  Ans.  12435  doUars. 

20.  In  one  mile  are  5280  feet ;  how  many  feet  in  8 
miles  ?  Ans.  42240  feet 

Case  IL 

63.  "When  the  miiltiplier  consists  of  two  or  more 
figures. 


1.  Multiply  746  by  23. 

OPERATION. 

746 


23 


Analysis.  Writ- 
ing the  multiplicand 
and  multiplier  as  in 
Case  I,  first  multiply 
each  figure  in  the 
multiplicand  by  the 
unit  figure  of  the 
multiplier,  precisely 
as  in  Case  I.  Then  multiply  by  the  2  tens.  2  tens  times  G  units,  or 
6  times  2  tens,  are  12  tens,  equal  to  1  hundred,  and  2  tens  ;  place  the 
2  tens  under  the  tens  figure  in  the  product  already  obtained,  and  add 
the  1  hundred  to  the  next  hundreds  produced.  2  tens  times  4  tens 
are  8  hundreds,  and  the  1  hundred  of  the  last  product  added  are  9 
hundreds  ;  write  the  9  in  hundreds'  place  in  the  product.    2  tens 


Multiplicand, 
Multiplier,    . 


Product, 


.  j  times  the  mul- 

\  tiplicand. 
nn  ]  time?  the  mul- 
•^^  j  tiplicand. 
go  j  times  the  mul- 

( tiplicand. 


Case  n  is  what  t    Give  explanation. 


40  SIMPLE    NUMBERS. 

times  7  hundreds  are  14  thousands,  equal  to  1  ten  thousand  and  4 
thousands,  which  we  write  in  their  appropriate  places  in  the  product. 
Then  adding  the  two  products,  the  entire  product  is  17158. 

1.  When  the  multiplier  contains  two  or  more  figures,  the  several  results  obtained 
by  multiplying  by  each  figure  are  called  particU  products. 

2.  When  there  are  ciphers  between  the  significant  figures  of  the  multiplier,  pass 
over  them,  and  multiply  by  the  significant  figures  only, 

63.  From  the  preceding  examples  and  illustrations  w€ 
deduce  the  following  general 

EiTLE.  I.  Write  the  multiplier  under  the  multiplicand  plac- 
ing units  of  the  same  order  under  each  other. 

II.  Multiply  the  multiplicand  ly  each  figure  of  the  multi- 
plier successively f  beginning  tuith  the  unitfigure,  and  torite 
the  first  figure  of  each  partial  product  imder  the  figure  of  the 
multiplier  used,  writing  down  and  carrying  as  in  addition. 

III.  If  there  are  partial  products y  add  them,  and  their  sum 
will  he  the  product  required, 

64.  Proof.  I.  Multiply  the  multiplier  by  the  multipli- 
cand, and  if  the  product  is.  the  same  as  the  first  result,  the 
work  is  correct.     Or, 

2.  Multiply  the  multiplicand  by  the  multiplier  diminished 
by  1,  and  to  the  product  add  the  multiplicand  ;  if  the  sum 
be  the  same  as  the  product  by  the  whole  of  the  multiplier, 
the  work  is  correct. 

Examples  for  Practice. 


Multiply 
By 

(2.) 

4732 
36 

(3.) 
8721 

47 

(4.) 
17605 
204 

28392 

61047 

70420 

14196 

34884 

35210 

Ans, 

170352 

409887 

3591420 

What  are  partial  products  ?  When  there  are  ciphers  in  the  multi- 
plier, how  proceed  ?  Rule,  first  step?  Second?  Third?  Proof,  first 
method?    Second? 


MULTIPLICATION-.  41 


(5.) 

(6.) 

(7.) 

7648 

81092 

37967 

328 

194 

426 

8.  How  many  yards  of  linen  in  759  pieces,  each  piece  con- 
taining 25  yards  ?  Ans.  18975  yards. 

9.  Sound  is  known  to  travel  about  1142  feet  in  a  second 
of  time  ;  how  far  will  it  travel  in  69  seconds  ? 

10.  A  9ian  bought  36  city  lots,  at  475  dollars  each ;  what 
did  they  all  cost  him?  Ans,  17100  dollars. 

11.  What  would  be  the  value  of  867  shares  of  railroad 
stock,  at  97  dollars  a  share  ?  Ans.  84099  dollars. 

12.  How  many  pages  in  3475  books,  if  there  be  362  pages 
in  each  book  ?  Ans.  1257950  pages. 

13.  In  a  garrison  of  4507  men,  each  man  receives  annu- 
ally 208  dollars  ;  what  do  they  all  receive  ? 

14.  Multiply  7198  by  216.  Ans,    1554768. 

15.  Multiply  31416  by  175.  Ans.    5497800. 

16.  Multiply  7071  by  556.  Ans.    3931476. 

17.  Multiply  75649  by  579.  Ans.  43800771. 

18.  Multiply  15607  by  3094.  Ans.  48288058. 

19.  Multiply  79094451  by  76095.  A7is.  6018692248845. 

20.  Multiply  five  hundred  forty  thousand  six  hundred 
nine,  by  seventeen  hundred  fifty.  Ans.  946065750. 

21.  Multiply  four  milUon  twenty-five  thousand  three 
hundred  ten,  by  seventy-five  thousand  forty-six. 

Ans.  302083414260. 

22.  Multiply  eight  hundred  seventy-seven  million  five 
hundred  ten  thousand  eight  hundred  sixty-four,  by  five 
hundred  forty-five  thousand  three  hundred  fifty-seven. 

Ans.   478556692258448. 

23.  If  one  mile  of  railroad  require  116  tons  of  iron,  worth 
65  dollars  a  ton,  what  will  be  the  cost  of  sufficient  iron  to 
construct  a  road  128  miles  in  length  ?    ' 

A71S.  965120  dollars. 


42  SIMPLE    l^UMBEKS, 

CONTEAOTIONS. 

Case  I. 

65.  "When  the  multiplier  is  a  composite  number. 

A  Composite  Number  is  one  tliat  may  be  produced  by 
multiplying  together  two  or  more  numbers ;  thus,  18  is  a 
composite  number,  since  6  x  3  =  18  ;  or,  9  x  2  =  18  ;  or, 
3x3x2  =  18. 

66.  The  Component  Factors  of  a  number  are  the  sev- 
eral numbers  which,  multiplied  together,  produce  the  given 
number ;  thus,  the  component  factors  of  20  are  10  and  2 
(10  X  2=20) ;  or,  4  and  5  (4  x  5=20)  ;  or,  2  and  2  and  5 
(2x2x5=20). 

The  pnpil  mnet  not  confbtmd  the  factors  with  the  parts  of  a  Dumber.  Thns,  the 
factors  of  which  13  is  composed,  are  4  and  3  (4  x  3=12) ;  while  the  parts  of  which  12 
38  composed  are  8  and  4  (8+4=12),  or  10  and  2  (10+2=12).  Tiie  factors  are  mutti- 
pliecl^  while  the  parts  are  added^  to  produce  the  number. 

1.  What  will  32  horses  cost,  at  174  doUars  apiece  ? 

OPERATION.  Analysis.    The  fac- 

Ifultlplicand,  174  COSt  of  1  horso.         tors  of  33  are  4  and 

l8t  factor,  4  ®-    If  we  multiply  the 


cost  of  1  horse  by  4, 
696  cost  of  4  horses.        ^e  obtain  the  cost  of  4 
9d  flMstor,  8  horses ;  and  by  multi- 

Product,  6568  cost  of  32  horses.     ?^^°^  *^ t  ^'*  ^^ .  ^ 

horses  by  8,  we  obtain 

the  cost  of  8  times  4  horses,  or  32  horses,  the  number  bought. 

6*7,  Hence  we  have  the  following 

KuLE.    I.  Separate  the  composite  numher  into  two  or  more 
factors. 
II.  Multiply  tlie  multiplicand  hy  one  of  these  factors,  and 

What  are  contractions  ?  Case  I  is  what  ?  Define  a  composite  num- 
ber. Component  factors.  What  caution  is  given  1  Give  explanation. 
Rule,  first  step  ?    Second  ? 


MULTIPLICATION.  43 

the  product  hy  another,  and  so  on  until  all  the  factors  have 
been  used  successively  j  the  last  product  will  be  the  product 
required. 

The  product  of  any  number  of  factors  wiU  be  the  same  in  whatever  order  they  are 
multiplied.    Thus,  4  x  3  >;  5=60,  and  5  x  4  x  3=  60. 

Examples  for  Practice. 

2.  Multiply  3472  by  48=6  x  8.  Ans,  166656. 

3.  Multiply  14761  by  64=8  x  8. 

4.  Multiply  87034  by  81=3  x  3  x  9.         Ans.  7049754. 

5.  Multiply  47326  by  120=6  x  5  x  4. 

6.  Multiply  60315  by  96.  Ans.  5790240. 

7.  Multiply  291042  by  125.  Ans.  36380250. 

8.  If  a  vessel  sail  436  miles  in  1  day,  bow  far  will  she  sail 
in  56  days  ?  A7is.  24416  miles. 

9.  What  will  72  acres  of  land  cost,  at  124  dollars  an  acre  ? 

Ans,  8928  dollars. 

10.  There  are  5280  feet  in  a  mile  ;  how  many  feet  in  84 
miles  ?  Ans.  443520  feet. 

11.  What  will  120  yoke  of  cattle  cost,  at  125  dollars  a 
yoke  ? 

Case  II. 

68.  "When  the  multiplier  is  10,  100,  1000,  etc. 

If  we  annex  a  cipher  to  the  multiplicand,  each  figure  is 
removed  one  place  toward  the  left,  and  consequently  the 
value  of  the  whole  number  is  increased  tenfold  ( 32 ).  If 
two  ciphers  are  annexed,  each  figure  is  removed  two  places 
toward  the  left,  and  the  value  of  the  number  is  increased 
one  hundred  fold ;  and  every  additional  cipher  increases 
the  value  tenfold. 

69.  Hence  the  following 

EuLE.  Annex  as  many  ciphers  to  the  multiplicand  as  there 
are  ciphers  in  the  multiplier  ;  the  number  so  formed  will  be 
the  product  required. 

Case  II  is  what  ?    Giv6  explanation.    Rule  2 


44  simple   kumbees. 

Examples  for  Practice. 

1.  Multiply  347  by  10.  Ans.  3470. 

2.  Multiply  4731  by  100.  Atis,  473100. 

3.  Multiply  13071  by  1000. 

4.  Multiply  89017  by  10000. 

5.  If  1  acre  of  land  cost  36  dollars,  what  will  10  acres 
cost  ?  Ans.  360  dollars^ 

6.  If  1  bushel  of  corn  cost  65  cents,  what  will  1000  bushels 
cost?  A?is,  65000  cents. 

Case  III. 

70.  "When  there  are  ciphers  at  the  right  hand  of 
one  or  both  of  the  factors. 

1.  Multiply  1200  by  60. 

OPERATION.  Analysis.    Both  multiplicand  and 

Multiplicand,  1200  multiplier  may  be  resolved  into  their 

Multiplier,  60  component  factors ;   1200  into  12  and 

— 100,  and  60  into  6  and  10.    If  these 

Product,  72000  several  factors  be  multiplied  together 

they  will  produce  the  same  product 
as  the  given  numbers  (67).  Thus,  12x6=7?,  and  72x100=7200, 
and  7200  x  10=72000,  vehich  is  the  same  result  as  in  the  operation. 

EuLE.  Multiply  the  significant  fig wes  of  the  multiplicand 
hy  those  of  the  multiplier,  and  to  the  product  annex  as  many 
ciphers  as  there  are  ciphers  on  the  right  of  both  factors. 

Examples  foe  Practice. 


Multiply 

By 

(2.) 
4720 
340000 

1888 
1416 

1604800000 

(3.) 

10340000 

105000 

5170 
1034 

1085700000000 

Cafle  III  is  ^'hat  ?    Give  explanation.    Rule. 


MULTIPLICATION,  45 

4.  Multiply  70340  by  800400.     Ans.  56300136000. 

5.  Multiply  3400900  by  207000.  Ans.  703986300000- 

6.  Multiply  634003000  by  40020.  Ans.  25372800060000. 

7.  Multiply  10203070  by  50302000. 

Ans.  513234827140000. 

8.  Multiply  30090800  by  600080.     Ans.  18056887264000. 

9.  Multiply  eighty  million  seven  thousand  six  hundred,  by 
eight  million  seven  hundred  sixty.    Ans.  640121605776000. 

10.  Multiply  fifty  million  ten  thousand  seventy,  by  sixty- 
four  thousand.  Ans.  3200644480000. 

11.  Multiply  ten  million  three  hundred  fifty  thousand  one 
hundred,  by  eighty  thousand  nine  hundred. 

Ans.  837323090000. 

12.  There  are  296  members  of  Congress,  and  each  one  re- 
ceives a  salary  of  3000  dollars  a  year ;  how  much  do  they 
all  receive  ? 

Examples  Combiking  Additioi^",  Subteactiok,  akd 
multiplicatiok. 

1.  Bought  45  cords  of  wood  at  4  dollars  a  cord,  and  9  loads 
of  hay  at  13  doUars  a  load  ;  what  was  the  cost  of  the  wood 
and  hay  ?  Ans.  297  dollars. 

2.  A  merchant  bought  6  hogsheads  of  sugar  at  31  dollars 
a  hogshead,  and  sold  it  for  39  dollars  a  hogshead  ;  what  did 
he  gain  ? 

3.  Bought  288  barrels  of  flour  for  1875  dollars,  and  sold 
the  same  for  9  dollars  a  barrel ;  what  was  the  gain  ? 

Ans.  717  dollars. 

4.  If  a  young  man  receive  500  dollars  a  year  salary  and 
pay  240  dollars  for  board,  125  dollars  for  clothing,  75  dollars 
for  books,  and  50  dollars  for  other  expenses,  how  much  will 
he  have  left  at  the  end  of  the  year  ?  Ans.  10  dollars. 

5.  A  farmer  sold  184  bushels  of  wheat  at  2  dollars  a 
bushel,  for  which  he  received  67  yards  of  cloth  at  4  dollars 
a  yard,  and  the  balance  in  groceries  ;  what  did  his  groceries 
cost  him  ? 


46  SIMPLE     N^UilBEBS. 

6.  A  sold  a  farm  of  320  acres  at  36  dollars  an  acre ;  B 
sold  one  of  244  acres  at  48  dollars  an  acre  ;  which  received 
the  greater  sum,  and  how  much?      Ans.  B,  192  dollars. 

7.  Two  persons  start  from  the  same  point  and  travel  in 
opposite  directions,  one  at  the  rate  of  35  miles  a  day,  and  the 
other  29  miles  a  day  :  how  far  apart  will  they  be  in  16  days  ? 

Ans,  1024  miles. 

8.  A  merchant  tailor  bought  14  bales  of  cloth,  each  bale 
containing  26  pieces,  and  each  piece  43  yards  ;  how  many 
yards  of  cloth  did  he  buy  ?  Ans.  15652  yards. 

9.  If  a  man  have  an  income  of  3700  dollars  a  year,  and 
his  daily  expenses  be  4  dollars  ;  what  will  he  save  in  a  year, 
or  365  days  ?  A7is.  2240  dollars. 

10.  A  man  sold  three  houses  ;  for  the  first  he  received 
2475  dollars,  for  the  second  840  dollars  less  than  he  received 
for  the  first,  and  for  the  third  as  much  as  for  the  olher  two  ; 
what  did  he  receive  for  the  three?         Ans.  8220  dollars. 

11.  A  man  sets  out  to  travel  from  Albany  to  Buffalo,  a 
distance  of  336  miles,  and  walks  28  miles  a  day  for  10  days  ; 
how  far  is  he  from  Buffalo  ? 

12.  Mr.  0  bo:ight  l4  cows  at  23  dollars  each,  7  horses  at 
96  dollars  each,  34  oxen  at  57  dollars  each,  and  300  sheep  at 
2  dollars  each ;  he  sold  the  whole  for  3842  dollars ;  what 
did  he  gain  ?  Ans.  310  dollars. 

13.  A  drover  bought  164  head  of  cattle  at  36  dollars  a 
head,  and  850  sheep  at  3  dollars  a  head ;  what  did  he  pay 
for  all  ? 

14.  A  banker  has  an  income  of  14760  dollars  a  year  ;  he 
pays  1575  dollars  for  house  rent,  and  four  tiihes  as  much  for 
family  expenses  ;  what  does  he  save  annually  ? 

Ans.  6885  dollars. 

15.  A  flour  merchant  bought  936  barrels  of  flour  at  9  dol- 
lars a  barrel ;  he  sold  480  barrels  at  10  dollars  a  barrel,  and 
the  remainder  at  8  dollars  a  barrel ;  what  did  he  gain  or 
lose  ?  Ans,  Gained  24  dollars. 


k 


DIVISIONe  4:t 

DIVISIOK 

Meistal  Exercises. 

71.  1.  How  many  hats,  at  4  dollars  apiece,  can  be  bought 

for  20  dollars  ? 

Analysis.  Since  4  dollars  will  buy  one  liat,  20  dollars  will  buy  as 
many  hats  as  4  is  contained  times  in  20,  which  is  5  times.  Therefore, 
5  hats,  at  4  dollars  apiece,  can  be  bought  for  20  dollars. 

2.  A  man  gave  16  dollars  for  8  barrels  of  apples ;  what 
was  the  cost  of  each  barrel  ? 

3.  If  1  cord  of  wood  cost  3  dollars,  how  many  cords  can 
be  bought  for  15  dollars  ? 

4.  At  6  shillings  a  bushel,  how  many  bushels  of  com  can 
be  bought  for  24  shilhngs  ? 

5.  When  flour  is  6  dollars  a  barrel,  how  many  barrels  can 
be  bought  for  30  dollars  ? 

6.  If  a  man  can  dig  7  rods  of  ditch  in  a  day,  how  many 
days  "v^U  it  take  him  to  dig  28  rods  ? 

7.  If  an  orchard  contain  56  trees,  and  7  trees  in  a  row, 
how  many  rows  are  there  ? 

8.  Bought  6  barrels  of  flour  for  42  dollars  ;  what  was  the 
cost  of  1  barrel  ? 

9.  If  a  farmer  divide  21  bushels  of  potatoes  equally 
among  7  laborers,  how  many  bushels  will  each  receive  ? 

10.  How  many  oranges  can  be  bought  for  27  cents,  at  3 
cents  each  ? 

11.  A  farmer  paid  35  dollars  for  sheep,  at  5  dollars  apiece  ; 
how  many  did  he  buy  ? 

12.  How  many  times  4  in  28  ?    In  16  ?    In  36  ? 

13.  How  many  times  8  in  40  ?     In  56  ?    In  64  ? 

14.  How  many  times  9  in  36  ?     In  63  ?    In  81  ? 

15.  How  many  times  7  in  49  ?    In  70  ?    In  84  ? 


48  SIMPLE     NUMBERS. 

73.  Division  is  the  process  of  finding  how  many  times 
one  number  is  contained  in  another. 

73.  The  Dividend  is  the  number  to  be  divided. 

74.  The  Divisor  is  the  number  to  divide  by. 

75.  The  Quotient  is  the  result  obtained  by  the  process 
of  division,  and  shows  how  many  times  the  divisor  is  con- 
tained in  the  dividend. 

1.  When  the  dividend  does  not  contain  the  divisor  an  exact  number  of  times,  the 
part  of  the  dividend  left  is  called  the  remainder^  and  it  must  be  less  than  the 
divisor. 

2.  As  the  remainer  is  always  a  part  of  the  dividend,  it  is  always  of  the  same  name 
or  kind. 

3.  When  there  is  no  remainder  the  division  is  said  to  be  exact. 

76.  The  sign,  -^,  placed  between  two  numbers,  denotes 
division,  and  shows  that  the  number  on  the  left  is  to  he  di- 
vided by  the  number  on  the  rigid.  Thus,  20-7-4=5,  is  read, 
20  divided  hy  4  is  equal  to  5. 

Division  is  also  indicated  by  writing  the  dividend  above, 

12 
and  the  divisor  helow  a  short  horizontal  line  ;  thus,  -q-=^> 

shows  that  12  divided  ly  3  equals  4. 

Case  I. 

77.  "When  the  divisor  consists  of  one  fignre. 
1.  How  many  times  is  4  contained  in  848  ? 

OPERATION.  Analysis.    After  writing  the  divisor 

T,,  , ,     ,  on  the  left  of  the  dividend,  with  a  line 

Dividend, 
.  .  ^  \  gig  between  them,  begin  at  the  left  hand 

and  say :  4  is  contained  in  8  hundreds, 

Qnotient,  212  3  hundreds  times,  and  write  2  in  hun- 

dreds' place  in  the  quotient ;  then  4  is 
contained  in  4  tens  1  ten  times,  and  write  the  1  in  tens'  place  in  the 
quotient ;  then  4  is  contained  in  8  units  2  units  times ;  and  writing  the 
2  in  units'  place  in  the  quotient,  and  the  entire  quotient  is  213. 

Define  division.  Dividend.  Divisor.  Quotient.  Remainder.  What 
is  complete  division ?  What  is  the  sign  of  division?  Case  I  is  what? 
Give  first  explanation. 


DIVISION.  49 

2.  How  many  times  is  4  contained  in  2884  ? 

OPERATION.  Analysis.    As  we  cannot  divide  2  thousands  by 

4  "i  2884  ^'  ^^®  *^^®  *^®  ^  thousands  and  the  8  hundreds  to- 

gether,  and  say,  4  is  contained  in  28  hundreds  7  hun- 

*^^  dreds  times,  which  we  write  in  hundreds'  place  in 

the  quotient ;  then  4  is  contained  in  8  tens  2  tens 
times,  which  we  write  in  tens'  place  in  the  quotient ;  and  4  is  con- 
tained in  4  units  1  unit  time,  which  we  write  in  units*  place  in  th^ 
quotient,  and  we  have  the  entire  quotient,  721. 

3.  How  many  times  is  6  contained  in  1824  ? 

OPERATION.  Analysis.    Beginning  as  in  the  last  example,  say, 

6  )  1824  6  is  contained  in  18  hundreds  3  hundreds  times, 

TTT  which  write  in  hundreds'  place  in  the  quotient ; 

then  6  is  contained  in  2  tens  no  times,  and  write  a 

cipher  in  tens*  place  in  the  quotient ;  and  taking  the  2  tens  and  4  units 

together,  6  is  contained  in  24  units  4  units  times,  which  write  in  units' 

place  in  the  quotient,  and  we  have  304  for  the  entire  quotient. 

4.  How  many  times  is  4  contained  in  943  ? 

OPERATION.  Analysis.    Here  4  is  contained  in  9 

^  \  9^3  hundreds  2  hundreds  times,  and  1  hundred 

over,  which,  united  to  the  4  tens,  makes 

235  ...  3  Eem.  ^4  ^^j^g  .  4  in  14  tens,  3  tens  times  and  3 
tens  over,  which,  united  to  the  3  units 
make  23  units ;  4  in  23  units  5  units  times  and  3  units  over.  The  3 
which  is  left  after  performing  the  division,  should  be  divided  by  4 ; 
but  the  method  of  doing  it  cannot  be  explained  until  we  reach 
Fractions  ;  so  we  merely  indicate  the  division  by  placing  the  divisor 
under  the  dividend,  thus,  f.  The  entire  quotient  is  written  235|, 
which  may  be  read,  two  hundred  thirty-five  and  three  divided  by  four ^ 
or,  two  hundred  thirty-five  and  a  remainder  of  three. 

From  the  foregoing  examples  and  illustrations,  we  deduce 
the  following 

KuLE.  I.  Write  the  divisor  at  the  left  of  the  dividend,  with 
a  line  between  them. 

Second.    Third.    Rule,  first  step  ? 
R.  P.  3 


60 


SIMPLE    NUMBERS. 


n.  Beginning  at  the  left  hand,  divide  each  figure  of  tlie 
dividend  hy  the  divisor,  and  write  the  result  binder  the  divi" 
dend, 

in.  If  there  he  a  remainder  after  dividing  any  figure,  re- 
gard it  as  prefixed  to  the  figure  of  the  next  loioer  order  in  the 
dividend,  and  divide  as  before. 

IV.  Should  any  figure  or  part  of  the  dividend  he  less  than 
the  divisor,  write  a  cipher  in  the  quotient,  and  prefix  the 
number  to  the  figure  of  the  next  lower  order  in  the  dividend, 
and  divide  as  before. 

V.  If  there  he  a  remainder  after  dividing  the  last  figure, 
place  it  over  the  divisor  at  the  right  hand  of  the  quotient. 

Proof.  Multiply  the  divisor  and  quotient  together,  and 
to  the  product  add  the  remainder,  if  any ;  if  the  result  be 
equal  to  the  dividend,  the  work  is  correct. 

1.  This  method  of  proof  depends  on  the  fact  that  division  is  the  reverse  of  mnlti- 
plication.  The  dividend  answers  to  the  product^  the  divisor  to  one  of  the  factors, 
and  the  quotient  to  the  other. 

2.  In  multiplication  the  two  factors  are  ariven,  to  find  the  product ;  in  division,  the 
product  and  one  of  the  £Eictors  are  given  to  find  the  other  £actor. 


Examples  for  Practice. 

1.  Divide  7824  by 

6. 

OPERATION. 

PROOF. 

Divisor.      6)7824 

Dividend. 

1304     Quotient. 

1304 

Quotient 

6      Divisor. 
7824     Dividend. 

(2.) 

(3.) 

(4.) 

4)65432 

5)89135 

6)178932 

(5.) 

(6.) 

(7.) 

7)4708935 

8)1462376 

9)7468542 

Second  step?    Tliird?    Fourth?    Fifth?    Proof?    How  does  divU 
ion  differ  from  multiplication  ? 


DIVISI02S". 


61 


8.  Divide  3102455  by  5. 

9.  Di^dde  1762891  by  4. 
19.  Divide  546215747  by  11. 
31.  Divide  30179624  by  12. 

12.  Divide  9254671  by  9. 

13.  Divide  7341568  by  7. 

14.  Divide  3179632  by  5. 
16.  Divide  19038716  by  8. 

16.  Divide  84201763  by  9. 

17.  Divide  2947691  by  12. 

18.  Divide  42084796  by  6. 

Sums  of  quotients  and  remainders, 


Quotients. 


Quotients. 

620491. 

440722f. 

49655977. 

2514968^. 

1028296|. 

Rem. 


20680083. 


28. 


19.  Divide  47645  dollars  equally  among  5  men ;  what  will 
each  receive?  Ans.  9529  dollars. 

20.  In  one  week  are  7  days ;  how  many  weeks  in  17675 
days  ?  A71S.  2525  weeks. 

21.  How  many  barrels  of  flour,  at  6  dollars  a  barrel,  can 
be  bought  for  6756  dollars?  Ans,  1126  barrels. 

22.  Twelve  things  make  a  dozen ;  how  many  dozen  in 
46216464  ?  Ans.  3851372  dozen. 

23.  How  many  barrels  of  flour  can  be  made  from  347560 
bushels  of  wheat,  if  it  take  5  bushels  to  make  one  barrel  ? 

A71S,  69512  barrels. 

24.  If  there  be  3240622  acres  of  land  in  11  townships, 
bow  many  acres  in  each  township  ? 

25.  A  gentleman  left  his  estate,  worth  38470  dollars,  to 
be  shared  equally  by  his  wife  and  4  children ;  what  did 
each  receive  ?  Ans.  7694  dollars. 

Case  IL 

78.  When  the  divisor  consists  of  two  or  more 
figures. 

To  illustrate  more  clearly  the  method  of  operation,  we  will  first  take  an  example 
usually  performed  by  Short  Division. 


Case  II  is  what  ? 


52  SIMPLE     i^UMBEKS. 

1.  How  many  times  is  8  contained  in  2528  ? 

OPERATION.  ANAi/f  SIS.    As  8  is  iiot  contained  in  2  thou- 

8)2528  (316  sands,  we  take  2  and  5  as  one  number,  and 

04.  consider  how  many  times  8  is  contained  in  tliis 

partial  dividend,  25  hundreds,  and  find  that  it 

1'®  is  contained  3  hundreds  times,  and  a  remainder. 

8  To  find  this  remainder,  we  multiply  the  divisor. 

An  8,  by  the  quotient  figure,  3  hundreds,  and  sub- 

tract the  product,  24  hundreds,  from  the  par- 

tial  dividend,  25  hundreds,  and  there  remains 

1  hundred.  To  this  remainder  we  bring  down 
the  2  tens  of  the  dividend,  and  consider  the  12  tens  a  second  partial 
dividend.  Then,  8  is  contained  in  12  tens  1  ten  time  and  a  remain- 
der ;  8  multiplied  by  1  ten  produces  8  tens,  which,  subtracted  from 
12  tens,  leave  4  tens.  To  this  remainder  we  bring  down  the  8  units, 
and  consider  the  48  units  the  third  partial  di\idend.  Then,  8  is  con- 
tained in  48  units  6  units  times.  Multiplying  and  subtracting  aa 
before,  we  find  that  nothing  remains,  and  the  entire  quotient  is  316. 

2.  How  many  times  is  23  contained  in  4807  ? 

OPERATION.  Analysis.    We  first  find  how 

Divisor.  Divid'd.  Quotient.  many  times  23  is  contained  in  48, 

23  )  4807  (  209  the  first  partial  dividend,  and  place 

46  the  result  in  the  quotient  on  the 

right  of  the  dividend.  We  then 
multiply  the  divisor,  23,  by  the 
quotient  figure,  2,  and  subtract  the 
product,  46,  from  the  part  of  the 
dividend  used,  and  to  the  remainder  bring  down  the  next  figure  of 
the  dividend,  which  is  0,  making  20,  for  the  second  partial  dividend. 
Then,  since  23  is  contained  in  20  no  times,  we  place  a  cipher  in  the 
quotient,  and  bring  down  the  next  figure  of  the  dividend,  making  a 
third  partial  dividend,  207  ;  23  is  contained  in  207,  9  times ;  multiply- 
ing and  subtracting  as  before,  nothing  remains,  and  the  entire  quo- 
tient is  209. 

1.  When  the  process  of  dividing  is  performed  mentally,  and  the  results  only  are 
written,  as  in  Case  I,  the  operation  is  termed  Short  Division. 

2.  When  the  whole  process  of  division  ia  written,  the  operation  is  termed  Long 
Division, 

Give  first  explanation.     Second.     What  is  long  division?    What  ia 
short  division  ?    When  is  each  used  ? 


207 
207 


DIVISIOK.  53 

8.  Short  Division  Is  generally  used  when  the  divisor  is  a  number  that  will  allow 
the  process  of  dividing  to  be  performed  mentally. 

From  the  preceding  illustrations  we  derive  the  following 
general 

KuLE.  I.  Write  the  divisor  at  the  left  of  the  dividend,  as 
in  short  division. 

II.  Divide  the  least  numher  of  the  left  hand  figures  in  tho 
dividend  that  will  contain  the  divisor  one  or  more  times,  and 
place  the  quotient  at  the  right  of  the  dividend,  with  a  line 
ietiueen  them, 

III.  Multiply  the  divisor  ly  this  quotient  figure,  suUract 
the  product  from  the  partial  dividend  used,  and  to  the  re- 
mainder hring  down  the  next  figure  of  the  dividend, 

IV.  Divide  as  hefore,  until  all  the  figures  of  the  dividend 
have  teen  Ir ought  down  and  divided, 

V.  If  any  partial  dividend  will  not  contain  the  divisor, 
place  a  cipher  in  the  quotient,  and  hring  down  the  next  figure 
of  the  dividend,  and  divide  as  hefore. 

YL  If  there  le  a  remainder  after  dividing  all  the  figures 
of  the  dividend,  it  7nust  be  written  in  the  quotient,  with  the 
divisor  underneath, 

1.  If  any  remainder  be  egval  to,  or  greater  than  the  divisor,  the  quotient  figure  is 
too  small,  and  must  be  increased. 

3.  If  the  product  of  the  divisor  by  the  quotient  figure  be  greoUr  than  the  pjirtial 
dividend,  the  quotient  figure  is  too  large,  and  must  be  diminished. 

79.  Peoof.  1.  The  same  as  in  short  division.     Or, 

2.  Subtract  the  remainder,  if  any,  from  the  dividend,  and 
di^ade  the  difference  by  the  quotient ;  if  the  result  be  the 
same  as  the  given  divisor,  the  work  is  correct. 

80.  The  operations  in  long  division  consist  of  five  prin^ 
cipal  steps,  viz. : 

1st.  Write  down  the  numbers. 

Rule,  first  step  ?  Second  ?  Third  ?  Fourth  ?  Fifth  ?  Sixth  ?  First 
direction  ?    Second  ?    Proof  ?     Recapitulate  the  steps  in  their  order. 


64  SIMPLE    KUMBEB  8. 

2d.    Find  how  many  times. 

3d.    Multiply. 

4th.  Subtract. 

5th.  Bring  down  another  figure. 

Examples  for  Practice. 
3.  Find  how  many  times  36  is  contained  in  11798. 


OPEKATION. 
Dividend. 
Divisor.     36  )  11798  (  327      Quotient. 
108 

PROOF  BY  MtJl 
327 

36 

[.TIPLICATIQH 

Quotient. 
Divisor. 

99 

72 

1962 
981 

278 
252 

11772 
26 

Remainder. 

26      Remainder. 

11798 

Dividend. 

4.  Find  how  many  times  82  ig 

!  contained  in  { 

B9634. 

OPERATION. 

B2  )  89634  (  1093 
82 

PROOF  BY  DIVISION. 
89634        Dividend. 
8        Remainder. 

763                               Quotient.       : 

738 

L093  )  89626  ( 
8744 

82       Divisor 

254 

246 

8 

2186 
2186 

5.  Find  how  many  times  154 

is  contained  in 

32740. 

6.  Divide  32572  by  34. 

7.  Divide  1554768  by  216. 

8.  Divide  5497800  by  175. 

9.  Divide  3931476  by  556. 
10.  Divide  10983588  by  132. 

Ans.  958. 
Ans,  7198. 
Ans,  31416. 
Ans.  7071. 
Ans.  83209. 

Divisioir.  55 

11.  Divide  73484248  by  19.  Ans.  3867592. 

12.  Divide  8121918  by  21.  Ans.    386758. 

13.  Divide  10557312  by  16.  Ans,    659832. 

14.  Divide  93840  by  63.  Ee?n,      33. 

15.  Divide  352417  by  29.  Bern.        9. 

16.  Divide  51846734  by  102.  Bern.      32, 

17.  Divide  1457924651  by  1204.  Bern.  1051 

18.  Divide  729386  by  731.  Bern.    679. 

19.  Divide  4843167  by  3605.  Bern.  1652. 

20.  Divide  49816657  by  9101.  •  Bern.  6884. 

21.  Divide  75867308  by  10115.  Bern,  4808. 

Quotients.  Rem. 

22.  Divide  28101418481  by  1107.  25385201.  974. 

23.  Divide  65358547823  by  2789.  23434402.  645. 

24.  Divide  102030405060  by  123456.  826451.  70404. 

25.  Divide  48659910  by  54001.  901.  5009. 

26.  Divide  2331883961  by  6739549.  346.  7. 

27.  A  railroad  cost  one  million  eigbt  hundred  fifty  thou- 
sand  four  hundred  dollars,  and  was  divided  into  eighteen 
thousand  five  hundred  and  four  shares  ;  what  was  the  value 
of  each  share  ?  Ans.  100  dollars. 

28.  If  a  tax  of  seventy-two  million  three  hundred  twenty 
thousand  sixty  dollars  be  equally  assessed  on  ten  thousand 
seven  hundred  thirty-five  towns,  what  amount  of  tax  must 
each  town  pay  ?  Ans.  6736^^^^  dollars. 

29.  In  1850  there  were  in  the  United  States  213  college 
Mbraries,  containing  942321  volumes ;  what  would  be  the 
average  number  of  volumes  to  each  library  ? 

Ans.  4424^  vols. 

30.  The  number  of  post  offices  in  the  United  States  in 
1853  was  22320,  and  the  entire  revenue  of  the  post  office 
department  was  5937120  doUars  ;  what  was  the  average 
revenue  of  each  office  ?  Ans,  266  dollars. 


56  SIMPLE    NUMBERS. 


CONTRACTIONS. 

Case  L 
81.  "When  the  divisor  is  a  composite  miniber. 

1.  If  3270  dollars  be  divided  equally  among  30  men,  ho-w 
many  dollars  will  each  receive  ? 

OPERATION.  Analysis.     If   8270   dollars   be   divided 

6  )  3270  equally  among  30  men,  each  man  will  receive 

^rr^77  as  many  dollars  as  30  is  contained  times  in 

I 3270  dollars.    30  may  be  resolved  into  the 

109  Ans,  factors  5  and  6 ;  and  we  may  suppose  the  30 
men  divided  into  5  groups  of  6  men  each; 
dividing  the  8270  dollars  by  5,  the  number  of  groups,  we  have  654, 
the  number  of  dollars  to  be  given  to  each  group ;  and  dividing  the 
654  dollars  by  6,  the  number  of  men  in  each  group,  we  have  109,  the 
number  of  dollars  that  each  man  will  receive. 

EuLE.  Divide  the  dividend  iy  one  of  the  factors,  and  the 
quotient  thus  obtained  hy  another,  and  so  on  if  there  be  more 
than  two  factors,  until  every  factor  has  been  made  a  divisor. 
The  last  quotient  will  be  the  quotient  required. 

Examples  for  Practice. 

2.  Divide  3690  by  15=3  x  5. 

3.  Divide  3528  by  24=4  x  6. 

4.  Divide  7280  by  35 = 5  x  7. 

5.  Divide  6228  by  36=6  x  6. 

6.  Divide  33642  by  27=3  x  9. 

7.  Divide  153160  by  56=7  x  8. 

8.  Divide  15625  by  125=5  x  5  x  5. 

83.  To  find  the  true  remainder. 

1.  Divide  1143  by  64,  using  the  factors  2,  8,  and  4,  antj 
find  the  true  remainder. 


Ans, 

246. 

Ans, 

147. 

Ans. 

208. 

Ans, 

173. 

Ans, 

1246. 

Ans, 

2735. 

Ans, 

125. 

What  are  contractions  1   Case  I  is  what  t   Give  explanation  ?   Rule. 


DITI8I0K.  67 

OPERATION.  Analysis.    Divid- 

2)1143  ing    1143    by  2,   we 

-TT—  haveaquotientofS?!, 

^1221 ^  ^^^  and  a  remainder  of 

4)  71 3x2=6     ''  1  undivided,  which, 

"l7. .  .3  X  8  X  2=48    <<  ^^^  I  ^.f'  f  *^^ 

—  given  dividend,  must 

55  true  rem.  ,  also  be  a  part  of  the 
true  remainder.  The 
671  being  a  quotient  arising  from  dividing  by  2,  its  units  are  2  times 
as  great  in  value  as  the  units  of  the  given  dividend,  1143.  Dividing 
the  571  by  8,  we  have  a  quotient  of  71,  and  a  remainder  of  8  undi- 
vided. As  this  3  is  a  part  of  the  571,  it  must  be  multiplied  by  2  to 
change  it  to  the  same  kind  of  units  as  the  1.  This  makes  a  true  re- 
mainder of  6  arising  from  dividing  by  8.  Dividing  the  71  by  4,  we 
have  a  quotient  of  17,  and  a  remainder  of  3  undivided.  This  3  is  a  part 
of  the  71,  the  units  of  which  are  S  times  as  great  in  value  as  those  of 
the  571,  and  the  units  of  the  5T.  i^re  2  times  as  great  in  value  as  those 
of  the  given  dividend,  1143  ;  therefove,  w  change  this  last  remainder, 
3,  to  units  of  the  same  value  as  the  dividend,  we  multiply  it  by  8  and 
2,  and  obtain  a  true  remainder  of  48  arising  from  dividing  by  4. 
Adding  the  three  partial  remainders,  we  obtain  55,  the  true  remainder. 

EuLE.  L  Multiply  each  partial  remainder,  except  the  fast, 
ly  all  the  preceding  divisors. 

II.  Add  the  several  products  with  the  first  remainder^  and 
the  sum  will  ie  the  true  remainder. 

Examples  for  Practice. 

Rem. 

2.  Divide        34712  by  42=6x7.  20, 

a  Divide      401376  by  64=8x8.  32. 

4.  Divide      139074  by  72=3x4x6.  42. 

5.  Divide    9078126  by  90=3  x  5  x  6.  6. 

6.  Divide  18730627  by  120=4  x  5  x  6.  67. 

7.  Divide    7360479  by  96=2x6x8.  63. 

8.  Divide  24726300  by  70=2  x  5  x  7.  60. 

9.  Divide    5610207  by  84=7  x  2  x  6.  15. 

Explain  the  process  of  finding  the  true  remainder  when  dividing  by 
Ihe  factors  of  a  composite  number. 

3* 


58  SIMPLE    ISrUMBERS. 

Case  IL 

83.  "When  the  divisor  is  10,  100,  1000,  etc. 

1.  Divide  374  acres  of  land  equally  among  10  men  ;  hew 
many  acres  will  each  have  ? 

OPEKATiON.  Analysis.    Since  we  have  shown, 

110)3714  ,  *^^*  ^  remove  a  figure  one  place 

■  toward  the  left  by  annexing  a  ciphei 

Quotient.     37 ...  4  Rem.  increases  its  value  tenfold,  or  multi. 

or,  373V  acres.  plies  it  by  10  ( 68 ),  so,  on  the  con- 

trary,  by  cutting  off  or  taking  away 

the  right  hand  figure  of  a  number,  each  of  the  other  figures  is  removed 

one  place  toward  the  right,  and,  consequently,  the  value  of  each  is 

diminished  tenfold,  or  divided  by  10  (  32 ). 

For  similar  reasons,  if  we  cut  ojff  two  figures,  we  divide  by 
100,  if  three,  we  divide  by  1000,  and  so  on. 

Rule.  From  the  right  hand  of  the  dividend  cut  off  as 
many  figures  as  there  are  ciphers  in  the  divisor.  Under  the 
figures  so  cut  off,  place  the  divisor,  and  the  whole  will  form 
the  quotient. 

Examples  for  Practice. 

2.  Divide  4760     by  10. 

3.  Divide  362078    by  100. 

4.  Divide  1306321   by  1000. 

5.  Divide  9760347   by  10000. 

6.  Divide  2037160310  by  100000. 

Case  III. 

84.  When  there  are  ciphers  on  the  right  hand  of 
the  divisor. 

I.  Divide  437661  by  800. 

OPERATION.  Analysis.     In  this  example  we 

8|00  )  4376|61  resolve  800  into  the  factors  8  and 


547... 61  Rem. 


100,  and  divide  first  by  100,  by  cut- 
ting off  two  right  hand  figures  of  the 


Case  II  is  what?     Give  explanation.     Rule.     Case  III  is  what? 
Give  explanation. 


DIVISION. 


S§ 


dividend  (  83 ),  and  we  have  a  quotient  of  4376,  and  k  remainder  of 
61.  We  next  divide  by  8,  and  obtain  547  for  a  quotient;  and  the 
entire  quotient  is  547/oV' 

2.  Dmde  34716  by  900. 

OPERATION.  Analysis.    Dividing  as 

9100  )  347116  ^  t^6  last  example,  we 

~  „  have  a  quotient  of  38,  and 

DO   Quotient,       O,  2d  rem,  ,  .    ,         ^r.        j  t 

^  '  two  remamders,  16  and  5 

5   X  100  +  16  =  516,  true  rem.  Multiplying  5,  the  last  re 

38f^,  Ans.  mainder,  by  100,  the  pre- 

ceding divisor,  and  adding 
16,  the  first  remainder  (82),  we  have  516  for  the  true  remainder. 
But  this  remainder  consists  of  the  last  remainder,  5,  prefixed  to  the 
figures  16,  cut  off  from  the  dividend. 

85.  When  there  is  a  remainder  after  dividing  by  the  sig- 
nificant figures,  it  must  be  prefixed  to  the  figures  cut  off 
from  the  dividend  to  give  the  true  remainder  ;  if  there  be 
no  other  remainder,  the  figures  cut  off  from  the  dividend 
wiU  be  the  true  remainder. 


Examples 

1  FOR  Peactice. 

Quotients.          Rem. 

3. 

Divide  34716 

by  900. 

38. 

516. 

4. 

Divide  1047634 

by  2400. 

436. 

1234. 

6. 

Divide  47321046 

by  45000. 

1051. 

26046. 

6. 

Divide  2037903176 

by  140000. 

63176. 

7. 

Divide  976031425 

by  92000. 

3425. 

8. 

Divide  80013176321 

by  700000. 

376321. 

9. 

Divide  19070367428 

by  4160000. 

4584. 

927428 

10. 

Divide  37902564431S 

►  by  554000000. 

89644319. 

11. 

The  circumference  of  the  earth  at  the 

equate 

)r  is  24898 

iles, 

.     How  many  hours  would  a  train  of  cars 

require  to 

travel  that  distance,  going  at  the  rate  of  50  miles  an  hour  ? 

A?is.  497|f. 
12.  The  sum  of  350000  dollars  is  paid  to  an  army  of  14000 


men  :  what  does  each  man  receive  ? 


Ans,  25  dollars. 


How  is  the  true  remainder  found  ? 


60  SIMPLE     N^UMBERS. 

Examples  iiq"  the  Preceding  Rules. 

1.  George  Washington  was  born  in  1732,  and  lived  67 
years ;  in  what  year  did  he  die  ?  A7is,  In  1799. 

2.  How  many  dollars  a  day  must  a  man  spend,  to  use  an 
income  of  1095  dollars  a  year?  A^is.  3  dollars. 

3.  If  I  give  141  dollars  for  a  piece  of  cloth  containing  47 
yards,  for  what  must  I  sell  it  in  order  to  gain  one  dollar  a 
yard?  Ans.  188  dollars. 

4.  A  speculator  who  owned  500  acres,  17  acres,  98  acres, 
and  121  acres  of  land,  sold  325  acres  ;  how  many  acres  had 
he  left  ?  Ans.  41 1  acres. 

5.  A  dealer  sold  a  cargo  of  salt  for  2300  dollars,  and  gained 
625  dollars  ;  what  did  the  cargo  cost  him  ? 

Ans.  1675  dollars. 

6.  If  a  man  earn  60  dollars  a  month,  and  spend  45  dol- 
lars in  the  same  time,  how  long  will  it  take  him  to  save  900 
dollars  from  his  earnings  ? 

7.  If  9  persons  use  a  barrel  of  flour  in  87  days,  how  many 
days  will  a  barrel  last  1  person  at  the  same  rate  ? 

A71S.  783  days. 

8.  The  first  of  three  numbers  is  4,  the  second  is  8  times 
the  first,  and  the  third  is  9  times  the  second  ;  what  is  their 
sum?  A71S.  324. 

9.  If  2,  2,  and  7  are  three  factors  of  364,  what  is  the 
other  factor  ?  Ans.  13. 

10.  A  man  has  3  farms ;  the  first  contains  78  acres,  the 
second  104  acres,  and  the  third  as  many  acres  as  both  the 
others  ;  how  many  acres  in  the  3  farms  ? 

11.  If  the  expenses  of  a  boy  at  school  are  90  dollars  for 
board,  30  dollars  for  clothes,  12  dollars  for  tuition,  5  dollars 
for  books,  and  7  dollars  for  pocket  money,  what  would  be  the 
expenses  of  27  boys  at  the  same  rate ?    A?is.  3888  dollars. 

12.  Four  children  inherited  2250  dollars  each  ;  but  one 
dying,  the  remaining  three  inherited  the  whole  ;  what  was 
the  share  of  each?  Ans.  3000  dollars. 


PKOMISCUOUS    EXAMPLES.  61 

13.  Two  men  travel  in  opposite  directions,  one  at  the  rate 
of  35  miles  a  day,  and  the  other  at  the  rate  of  40  miles  a 
day  ;  how  far  apart  are  they  at  the  end  of  '6  days  ? 

14.  Two  men  travel  in  the  same  direction,  one  at  the  rate 
of  35  miles  a  day,  and  the  other  at  the  rate  of  40  miles  a 
day  ;  how  far  apart  are  they  at  the  end  of  6  days  ? 

15.  A  man  was  45  years  old,  and  he  had  been  married  19 
years  ;  how  old  was  he  when  married  ?       Ans.  26  years. 

16.  Upon  how  many  acres  of  ground  can  the  entire  popu- 
lation of  the  globe  stand,  supposing  that  25000  persons  can 
stand  upon  one  acre,  and  that  the  population  is  1000000000  ? 

A?is.  40000  acres. 

17.  Add  384,  1562,  25,  and  946  ;  subtract  2723  from  the 
sum ;  divide  the  remainder  by  97  ;  and  multiply  the  quo- 
tient by  142  ;  what  is  the  result  ?  Ans.  284. 

18.  How  many  steps  of  3  feet  each  would  a  man  take  in 
walking  a  mile,  or  5280  feet  ?  Ans.  1760  steps. 

19.  A  man  purchased  a  house  for  2375  dollars,  and  ex- 
pended 340  dollars  in  repairs ;  he  then  sold  it  for  railroad 
stock  worth  867  dollars,  and  235  acres  of  western  land  val- 
ued at  8  dollars  an  acre  ;  what  did  he  gain  by  the  trade  ? 

Ans.  32  dollars. 

20.  The  salary  of  a  clergyman  is  800  dollars  a  year,  and 
his  yearly  expenses  are  450  dollars ;  if  he  be  worth  1350 
dollars  now,  in  how  many  years  will  he  be  worth  4500  dol- 
lars ?  Ans,  9  years. 

21.  How  many  bushels  of  oats  at  40  cents  a  bushel,  must 
be  given  for  1600  bushels  of  wheat  at  75  cents  a  bushel? 

A71S.  3000. bushels. 

22.  Bought  325  loads  of  wheat,  each  load  containing  50 
bushels,  at  2  dollars  a  bushel ;  what  did  the  wheat  cost  ? 

23.  If  you  deposit  225  cents  each  week  in  a  savings  bank, 
and  take  out  75  cents  a  week,  how  many  cents  will  you  have 
left  at  the  end  of  the  year  ?  Ans.  7800  cents. 

24.  The  product  of  two  numbers  is  31383450,  and  one  of 
the  numbers  is  4050  :  what  is  the  other  number? 


62  SIMPLE    NUMBERS. 

25.  The  Illinois  Central  Railroad  is  700  miles  long,  and 
cost  31647000  dollars  ;  what  did  it  cost  per  mile  ? 

A71S.  45210  dollars. 

26.  What  number  is  that,  which  being  divided  by  7,  the 
quotient  multiplied  by  3,  the  product  divided  by  5,  and  this 
quotient  increased  by  40,  the  sum  will  be  100  ?  Ans.  700. 

27.  How  many  cows  at  27  dollars  apiece,  must  be  given 
for  54  tons  of  hay  at  17  dollars  a  ton  ? 

28.  A  mechanic  receives  56  dollars  for  26  days'  work,  and 
spends  2  dollars  a  day  for  the  whole  time ;  how  many  dollars 
has  he  left  ?  Ans.  4:  dollars. 

29.  If  7  men  can  build  a  house  in  98  days,  how  long  would 
it  take  one  man  to  build  it  ?  Ans.  686  days. 

30.  The  number  of  school  houses  in  the  State  of  New 
York,  in  1855,  was  11,137;  if  their  cash  value  is  5,301,212 
dollars,  what  is  the  average  value  ?  A^is.  476  dollars. 

31.  A  cistern  whose  capacity  is  840  gallons  has  two  pipes ; 
through  one  pipe  60  gallons  run  into  it  in  an  hour,  and 
through  the  other  39  gallons  run  out  in  the  same  time ;  in 
how  many  hours  will  the  cistern  be  filled  ?    Ans.  40  hours. 

32.  The  average  beat  of  the  pulse  of  a  man  at  middle  age 
is  about  4500  times  in  an  hour ;  how  many  times  does  it 
beat  in  24  hours  ?  Ans.  108000  times. 

33.  How  many  years  from  the  discovery  of  America,  in 
1492,  to  the  year  1900? 

34.  According  to  the  census,  Maine  has  31766  square 
miles  ;  New  Hampshire,  9280 ;  Vermont,  10212  ;  Massachu- 
setts, 7800  ;  Rhode  Island,  1306  ;  Connecticut,  4674 ;  and 
New  York,  47000 ;  how  many  more  square  miles  has  all 
New  England  than  New  York  ? 

35.  What  is  the  remainder  after  dividing  62530000  by 
87900?  Ans.  33100. 

36.  A  pound  of  cotton  has  been  spun  into  a  thread  8  miles 
in  length  ;  allowing  235  pounds  for  waste,  how  many  pounds 
will  it  take  to  spin  a  thread  to  reach  round  the  earth,  suppos- 
ing the  distance  to  be  25000  miles?      Ans.  3360  pounds. 


PROMISCUOUS    EXAMPLES.  63 

37.  John  has  8546  dollars,  which  is  342  dollars  less  than 

4  times  as  much  as  Charles  has ;  how  many  dollars  has 
Charles?  Ans.  2222  dollars. 

38.  The  quotient  of  one  number  divided  by  another  is  37, 
the  divisor  245,  and  the  remainder  230  ;  what  is  the  divi- 
dend? Ans,  9295. 

39.  What  number  multiplied  by  72084  will  produce 
5190048?  Ans.  72. 

40.  There  are  two  numbers,  the  greater  of  which  is  73 
times  109,  and  their  difference  is  17  times  28  ;  what  is  the 
less  number  ?  Ans.  7481. 

41.  The  sum  of  two  numbers  is  360,  and  the  less  is  114 ; 
what  is  the  product  of  the  two  numbers  ?        Ans.  28044. 

42.  What  number  added  to  2473248  makes  2568754  ? 

43.  A  farmer  sold  35  bushels  of  wheat  at  2  dollars  a  bushel, 
and  18  cords  of  wood  at  3  dollars  a  cord  ;  he  received  9 
yards  of  cloth  at  4  dollars  a  yard,  and  the  balance  in  money  ; 
how  many  dollars  did  he  receive  ?  Ans.  88  dollars. 

44.  A  farmer  receives  684  dollars  a  year  for  produce  from 
his  farm,  and  his  expenses  are  375  dollars  a  year  ;  what  will 
he  save  in  five  years  ? 

45.  The  salt  manufacturer  at  Syracuse  pays  58  cents  for 
wood  to  boil  one  barrel  of  salt,  10  cents  for  boiling,  5  cents  to 
the  State  for  the  brine,  28  cents  for  the  barrel,  and  3  cents 
for  packing  and  weighing,  and  receives  125  cents  from  the 
purchaser  ;  what  does  he  make  on  a  barrel  ?    Ans.  21  cents. 

46.  A  company  of  15  persons  purchase  a  township  of 
western  land  for  286000  dollars,  of  which  sum  one  man  pays 
fjOOO  dollars,  and  the  others  the  remainder,  in  equal  amounts ; 
what  does  each  of  the  others  pay  ?        A7is.  20000  dollars. 

47.  If  256  be  multiplied  by  25,  the  product  diminished 
by  625,  and  the  remainder  divided  by  35,  what  will  be  the 
quotient?  Ans.  165. 

48.  Two  men  start  from  different  places,  distant  189  miles, 
and  travel  toward  each  other ;  one  goes  4  miles,  and  the  other 

5  miles  an  hour  ;  in  how  many  hours  will  they  meet  ? 


64  SIMPLE     NUMBEES. 

GENEEAL    PKINCIPLES    OF    DIVISION, 

86.  The  quotient  in  Division  depends  upon  the  relative 
values  of  the  dividend  and  divisor.  Hence  any  change  in  the 
value  of  either  dividend  or  divisor  must  produce  a  change 
in  the  value  of  the  quotient.  But  some  changes  may  be 
produced  upon  both  dividend  and  divisor,  at  the  same  time, 
that  will  not  affect  the  quotient.  The  laws  which  govern 
these  changes  are  called  Oeneral  Principles  of  Division, 
which  we  will  now  examine. 

I.  54  -T-  9  =  6. 

Multipljdng  the  dividend  by  3,  we  have 

54  X  3  -r-  9  =  162  -^  9  =  18, 

and  18  equals  the  quotient,  6,  multiplied  by  3.  Hence, 
Multiplying  the  dividend  ly  any  nuniber,  multiplies  the  quo- 
tient hy  the  same  number. 

II.  Using  the  same  example,  54  -h  9  =  6. 
Dividing  the  dividend  by  3  we  have 

■^-i-9=:18-^9  =  2, 

and  2  =  the  quotient,  6,  divided  by  3.  Hence,  Dividing  the 
dividend  ly  any  number,  divides  the  quotient  by  the  same 
number. 

III.  Multiplying  the  divisor  by  3,  we  have 

54  -^  9  X  3  =  54  -^  27  =  2, 

and  2  =  the  quotient,  6,  divided  by  3.  Hence,  Multiplying 
the  divisor  by  any  number,  divides  the  quotient  by  the  same 
mimber. 

IV.  Dividing  the  divisor  by  3,  we  have 

54  -J-  J  =  54  -f-  3  =  18, 

Upon  what  does  tlie  value  of  tlie  quotient  depend  ?    What  is  the 
first  general  principle  of  division  1    Second ?    Tliird  ?    Fourth? 


GEKERAL     PRINCIPLES    OF    DIVISION.       65 

ftnd  18  =  the  quotient,  6,  multiplied  by  3.     Hence,  Divid- 
ing the  divisor  hy  any  number,  multiplies  the  quotient  hy  the 
mme  number, 
V.  Multiplying  both  dividend  and  divisor  by  3,  we  have 
64  X  3  -V-  9  X  3  =  162  -^  27  =  6. 

Hence,  Multiplying  both  dividend  and  divisor  hy  the  same 
mmber,  does  not  alter  the  value  of  the  quotient, 

VL  Dividing  both  dividend  and  divisor  by  3,  we  have 

A^--|  =  18--3=:6. 

Pence,  Dividing  both  dividend  and  divisor  by  the  same  num* 
her,  does  not  alter  the  value  of  the  quotient. 

87.  These  six  examples  illustrate  all  the  different  changes 
we  ever  have  occasion  to  make  upon  the  dividend  and  divi- 
sor in  practical  arithmetic.  The  principles  upon  which 
these  changes  are  based  may  be  stated  as  follows  : 

Prin.  I.  Multiplying  the  dividend  multiplies  the  qicotient ; 
and  dividing  the  dividend  divides  the  quotient,  {SQ,  I  and  II.) 

Prin.  II.  Multiplying  the  divisor  divides  the  quotient; 
and  dividing  the  divisor  multiplies  the  quotient.  (86.  Ill 
and  IV.) 

Prin.  III.  Multiplying  or  dividing  both  dividend  and 
divisor  by  the  same  number,  does  not  alter  the  quotient,  (86. 
V  and  VL) 

88.  These  three  principles  may  be  embraced  in  one 

General  Law. 

A  change  i7i  the  dividend  produces  a  like  change  in  the 
quotient ;  but  a  change  in  the  divisor  produces  an  opposite 
change  in  the  quotient. 

I-  a  number  be  mnltiplied  and  the  product  divided  by  the  same  number,  the  quo- 
tient will  be  equal  to  the  number  multiplied.    Thus,  15  x  4=60,  and  60+4=15. 

Fifth?  Sixth?  Into  how  many  general  principles  can  these  br 
condensed?  What  is  the  first?  Second?  Third?  In  what  general 
law  are  these  embraced  ? 


66  PROPEETIES    OF    NUMBERS. 


EXACT    DIVISOES. 

89.  An  Exact  Divisor  of  a  number  is  one  that  gives 
a  whole  number  for  a  quotient. 

As  it  is  frequently  desirable  to  'know  if  a  number  has  an 
exact  divisor,  we  will  present  a  few  directions  that  will  be 
of  assistance,  particularly  in  finding  exact  divisors  of  large 
numbers. 

A  number  whose  unit  figure  is  0,  2,  4,  6,  or  8,  is  called  an  Even  Number.  And  a 
number  whose  unit  figure  is  1,  3,  5,  7,  or  9,  is  called  an  Odd  Number^ 

2  is  an  exact  divisor  of  all  even  numbers. 

4  is  an  exact  divisor  when  it  will  exactly  divide  the  tens 
and  units  of  a  number.  Thus,  4  is  an  exact  divisor  of  268, 
756,  1284. 

5  is  an  exact  divisor  of  every  number  whose  unit  figure  is 
0  or  5.     Thus,  5  is  an  exact  divisor  of  20,  955,  and  2840. 

8  is  an  exact  divisor  when  it  will  exactly  divide  the  hun- 
dreds, tens,  and  units  of  a  number.  Thus,  8  is  an  exact 
divisor  of  1728,  5280,  and  213560. 

9  is  an  exact  divisor  when  it  will  exactly  divide  the  sum  of 
the  digits  of  a  number.  Thus,  in  2486790,  the  sum  of  the 
digits  2  +  4  +  8  +  6  +  7  +  9  +  0=36,  and  36-t-9=4. 

10  is  an  exact  divisor  when  0  occupies  units'  place. 
100  when  00  occupy  the  places  of  units  and  tens. 

1000  when  000  occupy  the  places  of  units,  tens,  and  hun- 
dreds, etc. 

A  composite  number  is  an  exact  divisor  of  any  number, 
when  all  its  factors  are  exact  divisors  of  the  same  number. 
Thus,  2,  2,  and  3  are  exact  divisors  of  12  ;  and  so  also  are  4 
(=2x2)  and  6  (=2x3). 

An  even  number  is  not  an  exact  divisor  of  an  odd  number. 

If  an  odd  number  is  an  exact  divisor  of  an  even  number. 

What  is  an  exact  divisor  ?  What  is  an  even  number?  An  odd  num- 
ber? When  is  2  an  exact  divisor  ?  4?  5?  9?  10?  100?  1000? 
When  is  a  composite  number  an  exact  divisor  ?  An  even  number  is 
not  an  exact  divisor  of  what  ?  An  odd  number  is  an  exact  divisor  of 
what? 


FACTORIl^G    NUMBERS. 


67 


twice  that  odd  number  is  also  an  exact  divisor  of  the  even 
number.  Thus,  7  is  an  exact  divisor  of  42  ;  so  also  is  7  x  2, 
or  14. 

PRIME    NUMBEES. 

90.  A  Prime  K^umber  is  one  that  can  not  be  resolved 
or  separated  into  two  or  more  integral  factors. 

For  reference,  and  to  aid  in  determining  the  prime  factors 
of  composite  numbers,  we  give  the  following  : 

Table  of  Prime  Numbers  from  1  to  1000. 


1 

59 

139 

233 

337 

439 

557 

653 

769 

883 

2 

61 

149 

239 

347 

443 

563 

659 

773 

887 

3 

67 

151 

241 

349 

449 

569 

661 

787 

907 

5 

71 

157 

251 

353 

457 

571 

673 

797 

911 

7 

73 

163 

257 

359 

461 

577 

677 

809 

919 

11 

79 

167 

263 

367 

463 

587 

683 

811 

929 

13 

83 

173 

269 

373 

467 

593 

691 

821 

937 

17 

89 

179 

271 

379 

479 

599 

701 

823 

941 

19 

97 

181 

277 

383 

487 

601 

709 

827 

947 

23 

101 

191 

281 

389 

491 

607 

719 

829 

953 

29 

103 

193 

283 

397 

499 

613 

727 

839 

967 

31 

107 

197 

293 

401 

503 

617 

733 

853 

971 

37 

109 

199 

307 

409 

509 

619 

739 

857 

977 

41 

113 

211 

311 

419 

521 

631 

743 

859 

983 

43 

127 

223 

313 

421 

523 

641 

751 

863 

991 

47 

131 

227 

317 

431 

541 

643 

757 

877 

997 

53 

137 

229 

331 

433 

547 

647 

761 

881 

FACTORING    NUMBERS. 

Case  I. 

91.   To  resolve  any  composite  number  into  its 
prime  factors. 

What  is  a  prime  number  ?    In  factoring  numbers.  Case  I  is  what  ? 


2 

2772 

2 

1386 

3 

693 

3 

231 

7 

77 

11 

11 

1 

68  PROPERTIES    OF    NUMBERS. 

1  What  are  the  prime  factors  of  2772  ? 

OPERATION.  Analysis.    Divide  the  given  number  by  2, 

the  least  prime  factor,  and  the  result  by  2 ;  this 
gives  an  odd  number  for  a  quotient,  divisible  by 
the  prime  factor,  3,  and  the  quotient  resulting 
from  this  division  is  also  divisible  by  3.  The 
next  quotient,  77,  we  divide  by  its  least  prime 
factor,  7,  and  obtain  the  quotient  11 ;  this  being 
a  prime  number,  the  division  can  not  be  carried 
further.  The  divisors  and  last  quotient,  2,  2,  3, 
8,  7,  and  11  are  all  the  prime  factors  of  the  given 
number,  2772. 

Rule.  Divide  the  given  number  by  any  prime  factor ; 
divide  the  quotient  in  the  same  manner,  and  so  continue  the 
division  until  the  quotient  is  a  prime  number,  Tlie  several 
divisors  and  the  last  quotient  will  be  the  prime  factors  re- 
quired. 

Proof.  The  product  of  all  the  prime  factors  will  be  the 
given  number. 

Examples  for  Practice. 

2.  What  are  the  prime  factors  of  1140  ?    Ans,  2, 2, 3, 5, 19. 

3.  What  are  the  prime  factors  of  29925  ? 

4.  What  are  the  prime  factors  of  2431  ? 

5.  Find  the  prime  factors  of  12673. 

6.  Find  the  prime  factors  of  2310. 

7.  Find  the  prime  factors  of  2205. 

8.  What  are  the  prime  factors  of  13981  ? 

Case  II. 

92.  To  resolve  a  number  into  all  the  diflferent  sets 
of  factors  possible. 

1.  In  36  how  many  sets  of  factors,  and  what  are  they  ? 
Give  explanation.    Rule.    Proof.     Case  II  is  what  ? 


C  A  K  C  E  L  L  A  T  I  0  2^ 


69 


^2 

3 


OPERATION. 

xl8 
xl2 

4 

6 

2 
2 
3 

2 


36  =  ^ 


x9 
x6 
X  2  X 
x3  X 
X  3  X 
x2  X 


Analysis.  Writing  the  36  at 
tlie  left  of  the  sign  =,  arrange 
all  the  different  sets  of  factors  into 
which  it  can  be  resolved  under 
each  other,  as  shown  in  the  opera- 
tion, and  we  find  that  36  can  be 
resolved  into  8  sets  of  factors. 


x3 


Examples  for  Practice. 

2.  How  many  sets  of  factors  in  the  number  24  ?  "Wliat 
are  they?  Ans.  Q  sets. 

3.  In  125  how  many  sets  of  factors  ?    What  are  they  ? 

Ans,  2  sets. 

4.  In  40  how  many  sets  of  factors,  and  what  are  they  ? 

Ans.  6  sets. 

5.  In  72  how  many  sets  of  factors,  and  what  are  they  ? 

Ans.  15  sets. 

CANCELLATION. 

93.  Cancellation  is  the  process  of  rejecting  equal  fac- 
tors from  numbers  sustaining  to  each  other  the  relation  of 
dividend  and  divisor. 

It  has  been  shown  ( 7*7 )  that  the  dividend  is  equal  to  the 
product  of  the  divisor  multiplied  by  the  quotient.  Hence, 
if  the  dividend  can  be  resolved  into  two  factors,  one  of 
which  is  the  divisor,  the  other  factor  wiU  be  the  quotient. 

1.  Divide  63  by  7. 

Analysis.  We  see  in 
this  example  that  63  is 
composed  of  the  factors  7 
and  9,  and  that  the  factor 
7  is  equal  to  the  divisor. 

Therefore  we  reject  the  factor  7,  and  the  remaining  factor,  9,  is  the 

quotient. 


OPEKATION. 


Divisor,  ^  )  /^  X  9        Dividend. 

9        Quotient. 


Give  explanation.     What  is  cancellation  ?     Upon  what  principle  is 
it  based  ?    Give  first  explanation. 


70  PROPEETIES    OF    NUMBERS. 

94.  Whenever  the  dividend  and  divisor  are  each  com- 
posite numbers,  the  factors  common  to  both  may  first  be 
rejected  without  altering  the  final  result.     (  87,  Prin.  III.) 

2.  What  is  the  quotient  of  24  times  56  divided  by  7  times 
48? 

OPERATION.  Analysis.    First 

24x66_4x0x;?X$_.     J  indicate  the  opera- 

~~Z        77^  —  I        I        ^  —  ■*'  ^ns,      ^Jqjj  ^q  ^q  performed 

7  X  48  ^  X  0  X  ^  V       V     *i 

by  writing  the  num- 

<)ers  which  constitute  the  dividend  above  a  line,  and  those  which  con- 
Btitute  the  divisor  below  it.  Instead  of  multiplying  24  by  56,  in  the 
dividend,  we  resolve  24  into  the  factors  4  and  6,  and  56  into  the  factors 
7  and  8  ;  and  48  in  the  divisor  into  the  factors  6  and  8.  We  next 
cancel  the  factors  6,  7,  and  8,  which  are  common  to  the  dividend  and 
divisor,  and  we  have  left  the  factor  4  in  the  dividend,  which  is  the 
quotient. 

When  all  the  factors  or  numbers  in  tha  divid«ind  are  canceled,  1  should  be 
retained. 

95.  If  any  two  numbers,  one  in  the  dividend  and  one  in 
the  divisor,  contain  a  common  factor,  we  may  reject  that 
factor. 

3.  In  54  times  77,  how  many  times  63  ? 

OPERATION.  Analysis.    In  this  example  we  see  that  9  will 

Q        11  divide  54  and  63  :  so  we  reject  9  as  a  factor  of  54, 

^4-  \(  "^It  ^^^  retain  the  factor  6,  and  also  as  a  factor  of  63, 

and  retain  the  factor  7.     Again,  7  will  divide  7  in 

vP  the  divisor,  and  77  in  the  dividend.     Dividing 

'^  both  numbers  by  7,  1  will  be  retained  in  the 

divisor,  and   11  in  the  dividend.     Finally,  the 
product  of  6  X 11=:  66,  the  quotient. 

4.  Divide  25  x  16  x  12  by  10  x  4  x  6  x  7. 

OPERATION.  Analysis.    In  this, 

5          4^  as  in   the  preceding 

t$xHxn        5x4        ,,        ^,  example,  we  reject  all 

= =  *il^  =  -^-f •  the  factors  that   are 

40X^X0x7              7  common  to  both  divi- 

$  dend  and  divisor,  and 


Give  second  explanation. 


CANCELLATION.  71 

me  have  remaining  the  factor  7  in  the  divisor,  and  the  factors  5  and  4 
m  the  dividend.    Completing  the  work,  we  have  ^=2f,  Arts. 

From  the  preceding  examples  and  illustrations  we  derive 
the  following 

EuLE.  L  Write  the  numbers  composing  the  dividend  above 
a  horizontal  line,  and  the  numbers  composing  the  divisor 
below  it. 

II.  Cancel  all  the  factors  common  to  both  dividend  and 
divisor. 

III.  Divide  the  product  of  the  remaining  factors  of  the 
dividend  by  the  product  of  the  remaining  factors  of  the  di-^ 
visor,  and  the  result  will  be  the  quotient. 

1.  Kejecting  a  factor  from  any  number  is  dividing  the  number  by  that  &ctor. 

2.  When  a  factor  is  canceled,  the  unit,  1,  is  supposed  to  take  its  place. 

8.  One  factor  in  the  dividend  will  cancel  only  one  equal  factor  in  the  divisor. 

4.  If  all  the  factors  or  numbers  of  the  divisor  are  canceled,  the  product  of  the 
remaining  factors  of  the  dividend  will  be  the  quotient. 

5.  By  many  it  is  thought  more  convenient  to  write  the  fetors  of  the  dividend  on 
the  right  of  a  vertical  line,  and  tiie  fiactors  of  the  divisor  on  the  left, 

Examples  foe  Peactice, 
1.  What  is  the  quotient  of  16  x  5  x  4  divided  by  20  x  8  ? 

FIRST  OPERATION.  SECOND  OPERATION. 

.2 


2  . 

10X0X^       ^     ,  ^0 
=  2,  Ans,  rt 

A  


H' 
^ 


2,  Ans, 
2.  Divide  the  product  of  120  x  44  x  6  x  7  by  72  x  33  x  14. 

Rule,  first  step  ?  Second  ?  Third  ?  What  is  the  effect  of  rejecting 
a  factor  ?  What  is  the  quotient  when  all  the  factors  in  the  divisor  are 
canceled  \ 


n 


PEOPERTIES    OF    KHMBEKS, 
FIRST   OPERATION. 


10. 


r2 


xMx^xt      10x2 


nx$$xU 
^      3      ^ 


^  =  e^Ans. 


SECOND  OPERATION. 


10 


^ti 

m 

Ha 

0 

t 

3 

20 

6|, 

•  3.  Divide  the  product  of  33  >i:  35  x  28  by  11  x  15  x  14. 

Ans»  14. 

4.  What  is  the  quotient  of  21  x  11  X  26  divided  by  14  x 
13  ?  Ans.  33. 

5.  Divide  the  product  of  the  numbers  48,  72,  28,  and  5, 
by  the  product  of  the  numbers  84,  15,  7,  and  6,  and  give 
the  result.  Ans.  9-ij-. 

6.  Divide  140  x  39  x  13  x  7  by  30  x  7  x  26  x  21. 

Ans.  4J. 

7.  What  is  the  quotient  of  66  x  9  x  18  x  5  divided  by 
22  X  6  X  40.  ^  Ans.  lOJ. 

8.  Divide  the  product  of  200  x  36  x  30  x  21  by  270  x 
40  X  15  X  14.  Ans.  2, 

9.  Multiply  240  by  56,  and  divide  the  product  by  60  mul- 
tiplied by  28.  Ans.  8. 

10.  The  product  of  the  numbers  18,  6,  4,  and  42  is  to  be 
divided  by  the  product  of  the  numbers  4,  9,  3,  7  and  6  ; 
what  is  the  result  ?  Ans.  4. 

11.  How  many  tons  of  hay,  at  12  dollars  a  ton,  must  be 
given  for  30  cords  of  wood,  at  4  dollars  a  cord  ?  Ans.  10  tons. 


GREATEST    COMMON    DIVISOU.  73 

12.  How  many  firkins  of  butter,  each  containing  56 
pounds,  at  13  cents  a  pound,  must  be  given  for  4  barrels  of 
sugar,  each  containing  182  pounds,  at  6  cents  a  pound  ? 

Ans.  6  firkins. 

13.  A  tailor  bought  5  pieces  of  cloth,  each  piece  contain- 
ing 24  yards,  at  3  dollars  a  yard.  How  many  suits  of 
clothes,  at  18  dollars  a  suit,  must  be  m^de  from  the  cloth 
to  pay  for  it  ?  Ans.  20  suits. 

14.  How  many  days'  work,  at  75  cents  a  day,  will  pay  for 
115  bushels  of  corn,  at  50  cents  a  bushel  ?  Ans.  76f  days. 

GREATEST   COMMON  DIVISOR. 

96.  A  Common  Divisor  of  two  or  more  numbers  is  a 
number  that  will  exactly  divide  each  of  them. 

97.  The  Greatest  Common  Divisor  of  two  or  more 
numbers  is  the  greatest  number  that  will  exactly  divide 
each  of  them. 

Numbers  prime  to  each  other  are  such  as  have  no  com- 
mon di\dsor. 

A  common  diviBor  is  sometimes  called  a  Common  Measure ;  and  the  greatest 
common  divisor,  the  Greatest  Common 


Case  I. 
98.  "When  the  numbers  are  readily  factored. 

1.  What  is  the  greatest  common  divisor  of  6  and  10  ? 

Ans.  2. 

OPEBATION.  Analysis.    We  readily  find  by  inspection  tliat  2  will 

6  .  .  10        divide  both  tbe  given  numbers ;  hence  2  is  a  common 
"o         K        divisor ;  and  since  the  quotients  3  and  5  have  no  com- 

mon  factor,  but  are  prime  to  each  other,  the  common 

divisor,  2,  must  be  the  greatest  common  divisor. 

2.  What  is  the  greatest  common  divisor  of  42,  63,  and  105  ? 

What  is  a  common  divisor?  The  greatest  common  divisor?  A 
common  measure  ?  The  greatest  common  measure  ?  What  is  Case  I.? 
Give  analysis. 

'^^p,  4 


3 

42  . .  63 

.105 

7 

14  . .  21 

.  35 

2..  3. 

.   5 

74  PROPEBTIES    OF    UUMBBES. 

OPERATION.  ANAiiYSis.    We  observe  ttat  3  will 

exactly  divide  each  of  the  given  num- 
bers, and  that  7  will  exactly  divide 
each  of  the  resulting  quotients.  Hence, 
each  of  the  given  numbers  can  be  ex- 

3  ^  ij'_.21   Ans,  actly  divided  by  3  times  7  ;  and  these 

numbers  must  be  component  factors  of 
the  greatest  common  divisor.  Now,  if  there  were  any  other  component 
factor  of  the  greatest  common  divisor,  the  quotients,  2,  3,  5,  would  be 
exactly  divisible  by  it.  But  these  quotients  are  prime  to  each  other. 
Hence  3  and  7  are  all  the  component  factors  of  the  greatest  common 
divisor  sought. 

3.  What  is  the  greatest  common  divisor  of  28,  140,  and 
280? 

OPERATION.  Analysis.    We  first  divide  by  4  ; 

then  the  quotients  by  7.  The  re- 
sulting quotients,  1,  5,  and  10,  are 
prime  to  each  other.  Hence  4  and 
7  are  all  the  component  factors  of 

4  X  7=28,  A71S.  tt^6  greatest  common  divisor. 

From  these  examples  and  analyses  we  derive  the  following 

Rule.  I.  Write  the  numbers  in  a  line,  with  a  vertical  line 
at  the  left,  and  divide  ly  any  factor  common  to  all  the  numbers. 

XL  Divide  the  quotients  in  like  manner,  and  continue  the 
division  till  a  set  of  quotients  is  obtained  that  have  no  comm.on 
factor. 

III.  Multiply  all  the  divisors  together,  and  the  product  wiU 
be  the  greatest  common  divisor  sought. 

Examples  fob  Practice. 

1.  What  is  the  greatest  common  divisor  of  12, 36,  60,  72? 

Ans.  12. 

2.  What  is  the  greatest  common  divisor  of  18,  24,  30,  36, 
42?        ^  Ans,^. 

Rule,  first  step  ?    Second  ?    Third  ? 


4 

28  . .  140  . .  280 

7 

7..  35..  70 

1  . .   5  . ,  10 

i 


GREATEST    COMMON    DIVISOR.  75 

3.  What  is  the  greatest  common  divisor  of  72,  120,  240, 
384  ?  Ans.  24. 

4.  What  is  the  greatest  common  divisor  of  36,  126,  72, 
216  ?  Ans.  18. 

5.  What  is  the  greatest  common  divisor  of  42  and  112  ? 

Ans,  14. 

6.  What  is  the  greatest  common  divisor  of  32,  80,  and 
256?  Ans.ie, 

7.  What  is  the  greatest  common  divisor  of  210,  280,  350, 
630,  and  840  ?  Ans.  70. 

8.  What  is  the  greatest  common  divisor  of  300,  525,  225, 
and  375  ?  Ans,  75. 

9.  What  is  the  greatest  common  divisor  of  252,  630, 1134, 
and  1386?  .  A7is.  126. 

10.  What  is  the  greatest  common  divisor  of  96  and  544  ? 

Ans.  32. 

11.  What  is  the  greatest  common  divisor  of  468  and  1184  ? 

Ans.  4. 

12.  Wliat  is  the  greatest  common  divisor  of  200,  625,  and 
150  ?  Ans.  25. 

Case  II. 
99.  "When  the  mimbers  can  not  be  readily  factored. 

As  the  analysis  of  the  method  under  this  case  depends 
upon  three  properties  of  numbers  which  have  not  been  in- 
troduced, we  present  them  in  this  place. 

I.  An  exact  divisor  divides  any  number  of  times  its  divi- 
dend. 

II.  A  common  divisor  of  two  numbers  is  an  exact  divisor 
of  their  sum. 

III.  A  common  di^dsor  of  two  numbers  is  an  exact  divisor 
of  their  difference. 

What  is  Case  II.?  What  is  the  first  principle  upon  which  it  is 
founded?    Second?    Third? 


76 


PROPERTIES    OF    Is^  UMBERS. 


1.  What  is  the  greatest  common  divisor  of  84.  and  203  ? 

Analysis.  Draw  two  vertical  lines,  and 
place  the  larger  number  on  tlie  right,  and 
the  smaller  number  on  the  left,  one  line 
lower  down.  Then  divide  203,  the  larger 
number,  by  84,  the  smaller,  and  write  2, 
the  quotient,  between  the  verticals,  the 
product,  168,  opposite,  under  the  greater 
number,  and  the  remainder,  35,  below. 
Next  divide  84  by  this  remainder,  writing 
the  quotient,  2,  between  the  verticals,  the 
product,  70,  on  the  left,  and  the  new  remainder,  14,  below  the  70. 
Again  divide  the  last  divisor,  35,  by  14,  and  obtain  2  for  a  quotient, 
28  for  a  product,  and  7  for  a  remainder,  all  of  which  we  write  in  the 
game  order  as  in  the  former  steps.  Finally,  divide  the  last  divisor, 
14,  by  the  last  remainder,  7,  and  we  have  no  remainder.  7,  the  last 
divisor,  is  the  greatest  common  divisor  of  the  given  numbers. 


\JX 

-iSOVA 

203 

84 

2 

168 

70 

2 

35 

14 

2 

28 

14 

2 

7,  4'^s, 

0 

In  order  to  show  that  the  last  di^dsor  in  such  a  process  is 
the  greatest  common  divisor,  we  will  first  trace  the  work  in 
the  reverse  order,  as  indicated  by  the  arrow  line  below. 


OPERATION. 


84 


70 


14 


14 


7  divides  the  14,  as  proved  by 
the  last  division ;  it  will  also  di- 
vide two  times  14,  or  28,  (1.)  Now. 
as  7  divides  both  itself  and  28,  it 
will  divide  35,  their  sum,  (11.)  It 
will  also  divide  2  times  35,  or  70, 
(I ;)  and  since  it  is  a  common  di- 
visor of  70  and  14,  it  must  divide 
their  sum,  84,  which  is  one  of  the 
given  numbers,  (II.)  It  will  also 
divide  2  times  84,  or  168,  (I ;)  and 
since  it  is  a  common  divisor  of  108 
and  35,  it  must  divide  their  sum, 

203,  the  larger  number,  (II.)    Hence  7  is  a  common  divisor  of  the 

given  numbers. 
Again,  tracing  the  work  in  the  direct  order,  as  indicated  below,  we 


203 


168 


35 


28 


■^^ 


Give  analysis. 


GREATEST    COMMOK    DIVISOR. 


77 


know  that  the  greatest  common  divisor,  whatever  it  he,  must  divide 

2  times  84,  or    168,   (I.)      Then 


U203 


84   ^ 


70 


14 


168 


35 


28 


since  it  will  divide  both  168  and 
203,  it  must  divide  their  differ- 
ence, 35,  (III.)  It  will  also  divide 
2  times  35,  or  70,  (I ;)  and  as  it  will 
divide  both  70  and  84,  it  must  di- 
vide their  difference,  14,  (III.)  It 
will  also  divide  2  times  14  or  28, 
(I ;)  and  as  it  will  divide  both  28 
and  35,  it  must  divide  their  differ- 
ence, 7,  (III;)  hence,  it  canirwt  he 
greater  tJian  7 


Thus  we  haye  shown, 

1st.  That  7  is  a  common  divisor  of  the  given  numbers. 
2d.  That  their  greatest  common  divisor,  whatever  it  be, 
cannot  be  greater-  than  7.     Hence  it  must  be  7. 
From  this  example  and  analysis,  we  derive  the  following 

Rule.  I.  Draw  two  verticals,  and  write  the  two  numbers, 
one  on  each  side,  the  greater  7iuml)er  one  line  above  the  less. 

II.  Divide  the  greater  number  by  the  less,  tvriting  the  quo- 
tient between  the  verticals,  the  product  under  the  dividend, 
and  the  remainder  below, 

III.  Divide  the  less  number  by  the  remainder,  the  last  di' 
visor  by  the  last  remainder,  and  so  on,  till  nothing  remains. 
The  last  divisor  will  be  the  greatest  common  divisor  sought, 

IV.  If  more  than  tivo  numbers  be  given,  first  fi7id  the  great' 
est  common  divisor  of  two  of  them,  and  then  of  this  divisor 
and  one  of  the  remaining  numbers,  and  so  on  to  the  last; 
the  last  common  divisor  found  will  be  the  greatest  common 
divisor  of  all  the  given  numbers, 

1.  When  more  than  two  numbers  are  given,  it  is  better  to  begin  with  the  least  twe. 

2.  If  at  any  point  in  the  operation  a  prime  number  occur  as  a  remainder,  it  must 
be  a  common  divisor,  or  the  given  numbers  have  no  common  divisor. 


Rule,  first  step?    Second?    Third?    Fourth?    What  relation  have 
numbers  when  their  difference  is  a  prime  number  ? 


78 


PKOPEETIES    OF    NUMBERS. 


EXAMPf,ER 

FOR 

Practice. 

.  What  is  the  greatest  common  divisor  of  221  and  5512  ? 

OPERATION. 

5512 

221 

2 

442 

1092 

4 
1 

884 

208 

208 

Ans.  13 

1 

13 

78 

6 

78 

I        0 

2.  Find  the  greatest  common  divisor  of  154  and  210. 

Ans.  14. 

3.  What  is  the  greatest  common  divisor  of  316  and  664  ? 

Ans.  4. 

4.  What  is  the  greatest  common  divisor  of  679  and  1869  ? 

A}is.  7. 

5.  What  is  the  greatest  common  divisor  of  917  and  1495  ? 

Alls.  1. 

6.  What  is  the  greatest  common  divisor  of  1313  and  4108  ? 

Ans.  13. 

7.  What  is  the  greatest  common  divisor  of  1649  and  5423  ? 

Ans.  17. 

The  following  examples  may  be  solved  by  either  of  the 
foregoing  methods. 

8.  John  has  35  pennies,  and  Charles  50  :  how  shall  they 
arrange  them  in  parcels,  so  that  each  boy  shall  have  the  same 
number  in  each  parcel  ?  Ans.  6  in  each  parcel. 

9.  A  speculator  has  3  fields,  the  first  containing  18,  the 
second  24,  and  the  third  40  acres,  which  he  wishes  to  divide 
into  the  largest  possible  lots  having  the  same  number  of 
acres  in  each  ;  how  many  acres  in  each  lot?    Ans.  2  acres. 


MULTIPLES.  79 

10.  A  farmer  had  231  bushels  of  wheat,  and  273  bushels 
of  oats,  which  he  wished  to  put  into  the  least  number  of  bins 
containing  the  same  number  of  bushels,  without  mixing  the 
two  kinds  ;  what  number  of  bushels  must  each  bin  hold  ? 

Ans.  21. 

11.  A  Tillage  street  is  332  rods  long  ;  A  owns  124  rods 
front,  B  116  rods,  and  C  92  rods  ;  they  agree  to  divide  theii 
land  into  equal  lots  of  the  largest  size  that  will  allow  each 
one  to  form  an  exact  number  of  lots  ;  what  will  be  the  width 
of  the  lots?  Ans,  4  rods. 

12.  The  Erie  railroad  has  3  switches,  or  side  tracks,  of  the 
following  lengths  :  3013,  2231,  and  2047  feet ;  what  is  the 
length  of  the  longest  rail  that  vail  exactly  lay  the  track  on 
each  switch  ?  Ans.  23  feet. 

13.  A  forwarding  merchant  has  2722  bushels  of  wheat, 
1822  bushels  of  corn,  and  1226  bushels  of  beans,  which  he 
wishes  to  forward,  in  the  fewest  bags  of  equal  size  that  will 
exactly  hold  either  kind  of  grain ;  how  many  bags  will  it 
take  ?  Ans.  2885. 

14.  A  has  120  dollars,  B  240  dollars,  and  C  384  dollars ; 
they  agree  to  purchase  cows,  at  the  highest  price  per  head 
that  will  allow  each  man  to  invest  all  his  money  ;  how  many 
cows  can  each  man  purchase  ?    Ans.  A  5,  B  10,  and  C  16. 

MULTIPLES. 

100.  A  Multiple  is  a  number  exactly  divisible  by  a 
given  number  ;  thus,  20  is  a  multiple  of  4. 

101.  A  Common  Multiple  is  a  number  exactly  divisible 
by  two  or  more  given  numbers  ;  thus,  20  is  a  common  mul- 
tiple of  2,  4,  5,  and  10. 

102.  The  Least  Common  Multiple  is  the  least  num« 
ber  exactly  divisible  by  two  or  more  given  numbers  ;  thus, 
24  is  the  least  common  multiple  of  3,  4,  6,  and  8. 

What  is  a  multiple?  A  common  multiple?  The  least  commpp 
multiple? 


80  PROPERTIES     OF    JSTUAIBERS. 

103.  From  the  definition  ( 100 )  it  is  evident  that  the 
product  of  two  or  more  numbers,  or  any  number  of  times 
their  product,  must  be  a  common  multiple  of  the  numbers. 
Hence,  A  co7nmon  multiple  of  two  or  more  numbers  may  he 
found  ly  multiplying  the  given  numbers  together, 

104.  To  find  the  least  common  multiple. 

First  Method. 

From  the  nature  of  prime  numbers  we  derive  the  follow- 
ing principles  : 

I.  If  a  number  exactly  contain  another,  it  will  contain  all 
the  prime  factors  of  that  number. 

II.  If  a  number  exactly  contain  two  or  more  numbers,  it 
will  also  contain  all  the  prime  factors  of  those  numbers. 

III.  The  least  number  that  will  exactly  contain  all  the 
prime  factors  of  two  or  more  numbers,  is  the  least  common 
multiple  of  those  numbers. 

1.  Find  the  least  common  multiple  of  30,  42,  66,  and  78. 

OPERATION.  Analysis.      The 

30  =  2  X  3  x5  number  cannot  fee 

42  =  2  X  3  X  7  ^®^®  than  78,  since 

aa        o  v^  Q  v^  11  it  must  contain  78 ; 

bb  =  /i  X  o  X  11  ,  .^         ^ 

7Q  _  9   ^  q  ^  iq  ^e^ce  It  must  con. 

2x3x13x11x7x5  =  30030,  A7is.        78,  viz. : 

2  X  3  X  13. 
We  here  have  all  the  prime  factors  of  78,  and  also  all  the  factors  ol 
66,  except  the  factor  11.     Annexing  11  to  the  series  of  factors, 
2  X  3  X  13  X  11, 

and  we  have  all  the  prime  factors  of  78  and  66,  and  also  all  the  factors 
of  42  except  the  factor  7.     Annexing  7  to  the  series  of  factors, 

2  X  3  X  13  X  11  X  7, 

and  we  have  all  the  prime  factors  of  78,  66,  and  42,  and  also  all  the 

How  can  a  common  multiple  of  two  or  more  numbers  be  found  ? 
First  principle  derived  from  prime  numbers  ?  Second  ?  Third  % 
Give  anal)'sis. 


LEAST    COMMOK    MULTIPLE.  81 

factors  of  30  except  the  factor  5.     Annexing  5  to  the  series  of  factors, 

3  X  3  X  13  X  11  X  7  X  5, 

and  we  have  all  the  prime  factors  of  each  of  the  given  numbers  ;  and 
hence  the  product  of  the  series  of  factors  is  a  common  multiple  of  the 
given  numbers,  (II.)  And  as  no  factor  of  this  series  can  be  omitted 
without  omitting  a  factor  of  one  of  the  given  numbers,  the  product  of 
the  series  is  the  least  common  multiple  of  the  given  numbers,  (III.) 

From  this  example  and  analysis  we  deduce  the  following 

Rule.  I.  Resolve  the  given  numbers  into  their  prime 
factors. 

II.  Take  all  the  prime  factors  of  the  largest  number,  and 
such  prime  factors  of  the  other  numbers  as  are  not  found  in 
the  largest  number,  and  their  product  will  be  the  least  com- 
mon multiple. 

When  a  prime  factor  is  repeated  in  any  of  the  given  numbers,  it  must  be  used  as 
many  times,  as  a  factor  of  the  multiple,  as  the  greatest  number  of  times  it  appears 
iu  any  of  the  given  numbers. 


Examples  for  Practice. 

2.  Find  the  least  common  multiple  of  7,  35,  and  98. 

Ans.  490. 

3.  Find  the  least  common  multiple  of  24,  42,  and  17. 

Ans.  2856. 

4.  What  is  the  least  common  multiple  of  4,  9,  6,  8  ? 

Ans.  72. 
6.  What  is  the  least  common  multiple  of  8,  15,  77,  385  ? 

Ans.  9240. 

6.  What  is  the  least  common  multiple  of  10,  45,  75,  90  ? 

Ans.  450. 

7.  Wliat  is  the  least  common  multiple  of  12,  15,  18,  35  ? 

Ans.  1260. 

Rule,  first  step  ?    Second  ?    What  caution  is  given  ? 
4* 


2 

OPERATION. 
4.  .6.  .9. 

12 

2 

2 

..3. .9. 

.  6 

3 

3. .9. 

.  3 

3 

3 

82  properties   of   numbers. 

Secoi^d  Method. 

105.    1.  What  is  the  least  common  multiple  of  4,  6,  9, 
and  12? 

ANAiiYSis.  First  write  the 
given  numbers  in  a  series,  with 
a  vertical  line  at  the  left 
Since  2  is  a  factor  of  some  oi 
the  given  numbers,  it  must  be 
a  factor  of  the  least  common 
multiple  sought.  Dividing  as 
2x2x3x3  =  36,  Ans,  many  of  the  numbers  as  are 

divisible  by  2,  write  the  quo- 
tients and  the  undivided  number,  9,  in  a  line  underneath.  We  no\« 
perceive  that  some  of  the  numbers  in  the  second  line  contain  the 
factor  2 ;  hence  the  least  common  multiple  must  contain  another  2, 
and  we  again  divide  by  2,  omitting  to  write  down  any  quotient  when 
it  is  1.  We  next  divide  by  3  for  a  like  reason,  and  still  again  by  3. 
By  this  process  we  have  transferred  all  the  factors  of  each  of  the 
numbers  to  the  left  of  the  vertical ;  and  their  product,  36,  must  be  the 
least  common  multiple  sought,  (104,  III.) 

2.  What  is  the  least  common  multiple  of  10, 12, 15,  and  75  ? 

Analysis.  We  read- 
ily see  that  2  and  5  are 
among  the  factors  of  the 
given  numbers,  and  must 
be  factors  of   the  least 

2x5x2x3x5=  300,  Ans.       ^"^""^^  "'^^^^P^^ '  ^'"'^ 

we  divide  every  number 
that  is  divisible  by  either  of  these  factors  or  by  their  product ;  thus,  we 
divide  10  by  both  2  and  5  ;  12  by  2  ;  15  by  5  ;  and  75  by  5.  We  next 
divide  the  second  line  in  like  manner  by  2  and  3 ;  and  afterwards  the 
third  line  by  5.  By  this  process  we  collect  the  factors  of  the  given 
numbers  into  groups;  and  the  product  of  the  factors  at  the  left  of  the 
vertical  is  the  least  common  multiple  sought. 

3.  Wliat  is  the  least  common  multiple  of  6,  15,  35,  42, 
and  70  ? 


2,5 

OPERATION. 

10  . .  12  . .  15  . .  75 

2,3 

6..  3..  15 

5 

5 

Give  explanation. 


3,7 

2,5 


LEAST    COMMOIT    MULTIPLE.  OO 

OPEBATiON.  Analysis.    In  this  opera- 

15  .  .  42  .  .  70  tion  we  omit  the  6  and  35, 

because  they  are  exactly  con- 
tained in  some  of  the  other 


5  .  .    2  . .  10 


3x7x2x5  =  210,  Ans,  given    numbers ;    thus,    6  is 

contained  in  43,  and  35  in 
70;  and  whatever  will  contain  42  and  70  must  contain  6  and  35. 
Hence  we  have  only  to  find  the  least  common  multiple  of  the  remain- 
ing numbers,  15,  42,  and  70. 

From  these  examples  we  derive  the  following 

KuLE.  I.   Write  the  numbers  in  a  line,  omitting  any  of  the 

smaller  numbers  that  are  factors  of  the  larger,  and  draw  a 

vertical  line  at  the  left. 

II.  Divide  hy  any  prime  factor,  or  factors,  that  may  he  con- 
tained in  one  or  more  of  the  given  numbers,  and  write  the 
quotients  and  undivided  numbers  i7i  a  line  underneath,  omit- 
ting the  Vs. 

III.  In  nice  manner  divide  the  quotients  and  undivided 
numbers,  and  continue  the  process  till  all  the  factors  of  the 
given  numbers  have  been  transferred  to  the  left  of  the  vertical. 
Tfien  multiply  these  factors  together,  and  their  product  will 
be  the  least  common  multiple  required. 

Examples  for  Practice. 

4.  What  is  the  least  common  multiple  of  12,  15,  42,  and 
60?  Ans.  420. 

5.  What  is  the  least  common  multiple  of  21,  35,  and  42  ? 

Ans.  210. 

6.  What  is  the  least  common  multiple  of  25,  60, 100,  and 
125?  Ans.  1500. 

7.  What  is  the  least  common  multiple  of  16,  40,  96,  and 
105?  Ans.  3360. 

8.  What  is  the  least  common  multiple  of  4, 16,  20,  48,  60, 
and  72  ?  Ans.  720. 

9.  What  is  the  least  common  multiple  of  84, 100,  224,  and 
300?  Ans.  16800. 

Rule,  first  step  ?    Second  ?    Third  ? 


84  PROPERTIES     OF    NUMBERS. 

10.  What  is  the  least  common  multiple  of  270,  189,  297, 
243?  Ans.  187110. 

11.  What  is  the  least  common  multiple  of  1,  2,  3,  4,  5,  6, 
7,  8,  9  ?  Ans.  2520. 

12.  What  is  the  smallest  sum  of  money  for  which  I  could 
purchase  an  exact  number  of  books,  at  5  dollars,  or  3  dol- 
lars, or  4  dollars,  or  6  dollars  each  ?  Ans,  60  dollars. 

13.  A  farmer  has  3  teams  ;  the  first  can  draw  12  barrels 
of  flour,  the  second  15  barrels,  and  the  third  18  barrels ; 
what  is  the  smallest  number  of  barrels  that  will  make  full 
loads  for  any  of  the  teams  ?  Ans,  180. 

14.  What  is  the  smallest  sum  of  money  with  which  I  can 
purchase  cows  at  130  each,  oxen  at  $55  each,  or  horses  at 
$105  each?  Ans.  U310, 

15.  A  can  shear  41  sheep  in  a  day,  B  63,  and  C  54  ;  what 
is  the  number  of  sheep  in  the  smallest  flock  that  would 
furnish  exact  days'  labor  for  each  of  them  shearing  alone  ? 

Ans,  15498. 

16.  A  servant  being  ordered  to  lay  out  equal  sums  in  the 
purchase  of  chickens,  ducks,  and  turkeys,  and  to  expend  as 
little  money  as  possible,  agreed  to  forfeit  5  cents  for  every 
fowl  purchased  more  than  was  necessary  to  obey  orders.  In 
the  market  he  found  chickens  at  12  cents,  ducks  at  30  cents, 
and  turkeys  at  two  prices,  75  cents  and  90  cents,  of  which 
he  imprudently  took  the  cheaper ;  how  much  did  he  thereby 
forfeit  ?  Ans.  80  cents. 


CLASSIFICATION    OF    NUMBEES. 
Numbers  may  be  classified  as  follows  : 

106.  I.  As  Even  and  Odd. 

107.  II.  As  Prime  and  Composite. 

What  is  the  first  classification  of  numbers  ?  What  is  an  even  num- 
ber? An  odd  number?  Second  classification?  A  prime  number? 
A  composite  number  ? 


.CLASSIFICATION    OF    KUMBERS.  85 

108.  III.  As  Integral  and  Fractional 

An  Integral  Number,  or  Integer,  expresses  whole 
things.     Thus,  281 ;  78  boys;  1000  books. 

A  Fractional  Number,  or  Fraction,  expresses  equal 
parts  of  a  thing.  Thus,  half  a  dollar  ;  three-fourths  of  an 
hour ;  seven-eighths  of  a  mile. 

109.  IV.  As  Abstract  and  Concrete, 

110.  V.  As  Simple  and  Compound. 

A  Simple  Number  is  either  an  abstract  number,  or  a 
concrete  number  of  but  one  denomination.  Thus  48,  926 ; 
48  dollars,  926  miles. 

A  Compound  Number  is  a  concrete  number  whose  value 
is  expressed  in  two  or  more  different  denominations.  Thus, 
32  dollars  15  cents;  15  days  4  hours  25  minutes;  7  miles 
82  rods  9  feet  6  inches. 

111.  YI.  As  Lihe  and  Unlike, 

Like  Numbers  are  numbers  of  the  same  unit  value. 

If  simple  numbers,  they  must  be  all  abstract,  as  6,  62, 
487 ;  or  all  of  one  and  the  same  denomination,  as  5  apples, 
62  apples,  487  apples ;  and, ,  if  compound  numbers,  they 
must  be  used  to  express  the  same  kind  of  quantity,  as  time, 
distance,  etc.  Thus,  4  weeks  3  days  16  hours;  1  week  6 
days  9  hours ;  5  miles  40  rods ;  2  miles  100  rods. 

Unlike  Numbers  are  numbers  of  different  unit  values. 
Thus,  75,  140  dollars,  and  28  miles ;  4  hours  30  minutes, 
and  5  bushels  1  peck.  / 

What  is  the  third  classification  ?  What  is  an  integral  number  ?  A 
fractional  number  ?  What  is  the  fourth  classification  ?  An  abstract 
number  ?  A  concrete  number  ?  What  is  the  fifth  classification  ?  A 
simple  number  ?  A  compound  number  ?  Sixth  classification  ?  What 
are  like  numbers  ?    UnliJie  numbers  ? 


86 


FRACTIOifS. 


FEACTIOl^S. 

Definitions,  Notation,  and  Numeration. 

112.  If  a  unit  be  divided  into  2  equal  parts,  one  of  the 
parts  is  called  one-half. 

If  a  unit  be  divided  into  3  equal  parts,  one  of  the  parts  is 
called  one-third,  two  of  the  parts  two-thirds. 

If  a  unit  be  divided  into  4  equal  parts,  one  of  the  parts 
is  called  one-fourth,  two  of  the  parts  two-fourths,  three  of 
the  parts  three-fourths. 

If  a  unit  be  divided  into  5  equal  parts,  one  of  the  parts 
is  called  one-fifth,  two  of  the  parts  two-fifths,  three  of  the 
parts  three-fifths,  etc. 

The  parts  are  expressed  by  figures ;  thus. 


One-half  is  written 

i 

One-fifth    is  written 

i 

One-third           " 

i 

Two-fifths 

1 

Two-thirds        " 

f 

One-seventh    ^    '' 

+ 

One-fourth        " 

i 

Three-eighths     '* 

1 

Two-fourths       " 

f 

Five-ninths          " 

1 

Three-fourths    " 

f 

Eight-tenths        " 

A 

Hence  we  see  that  the  parts  into  which  a  unit  is  divided 
take  their  name,  and  their  value,  from  the  number  of  equal 
parts  into  which  the  unit  is  divided.  Thus,  if  we  divide 
an  orange  into  2  equal  parts,  the  parts  are  called  halves  ; 
if  into  3  equal  parts,  thirds  ;  if  into  4  equal  parts,  fourths, 
etc. ;  and  each  third  is  less  in  value  than  each  half,  and 
esLoh  fourth  less  than  each  third;  and  the  greater  the  num- 
ber of  parts,  the  less  their  value. 

When  a  unit  is  divided  into  any  number  of  equal  parts, 
one  or  more  such  parts  is  a  fractional  part  of  the  whole 
number,  and  is  called  a.  fraction.     Hence, 

113.  A  Fraction  is  one  or  more  of  the  equal  parts  of  a 
unit. 


Define  a  fraction. 


DEFINITIONS,    NOTATION,    NUMERATION.       87 

114.  To  write  a  fraction,  two  integers  are  required,  one 
to  express  the  number  of  parts  into  which  the  whole  num- 
ber is  divided,  and  the  other  to  express  the  number  of  these 
parts  taken.  Thus,  if  one  dollar  be  divided  into  4  equal 
parts,  the  parts  are  called  fourths,  and  three  of  these  parts 
are  called  three-fourths  of  a  dollar.  This  three-fourths  may 
be  written 

3  the  number  of  parts  taken. 

4  the  number  of  parts  into  which  the  dollar  is  divided. 

115.  The  Denominator  is  the  number  below  the  line. 
It  denominates  or  names  the  parts ;  and 

It  shows  how  many  parts  are  equal  to  a  unit. 

116.  The  iN'umerator  is  the  number  above  the  line. 
It  numerates  or  numbers  the  parts ;  and 

It  shows  how  many  parts  are  taken  or  expressed  by  the 
fraction. 

117.  The  Terms  of  a  fraction  are  the  numerator  and 
denominator,  taken  together. 

118.  Fractions  indicate  division,  the  numerator  answer- 
ing to  the  dividend,  and  the  denominator  to  the  divisor. 

110.  The  Value  of  a  fraction  is  the  quotient  of  the  nu- 
merator divided  by  the  denominator. 

120.  To  analyze  a  fraction  is  to  designate  and  describe 
its  numerator  and  denominator.  Thus,  j  is  analyzed  as 
follows  : — 

4  is  the  denomin^itor,  and  shows  that  the  integer  is  divided 
into  4  equal  parts ;  it  is  the  divisor. 

3  is  the  numerator,  and  shows  that  3  parts  are  taken ;  it 
is  the  dividend,  or  integer  divided. 

3  and  4  are  the  ter7ns,  considered  as  dividend  and  divisor. 

The  value  of  the  fraction  is  the  quotient  of  3-i-4,  or  |. 

HoTV  many  numbers  are  required  to  write  a  fraction  ?  Why  ?  De- 
fine the  denominator.  The  numerator.  What  are  the  terms  of  a  frac- 
tion ?    The  value  ?    Whf>t  is  the  analysis  of  a  fraction  ? 


88  feacixons. 

Examples  fok  Practice. 

Express  the  following  fractions  by  figures  : 

1.  Seven  eighths, 

2.  Three  twenty-fifths. 

3.  Nine  one-hundredths. 

4.  Sixteen  thirtieths. 

5.  Thirty-one  one  hundred  eighteenths. 

6.  Seventy-five  ninety-sixths. 

7.  Two  hundred  fifty-four /owr  hundred  forty-thirds. 

8.  Eight  nine  hundred  twenty-firsts. 

9.  One  thousand  two  hundred  thirty-two  seventy-five  thot^ 
sand  six  hundredths. 

10.  Nine  hundred  six  two  hundred  forty-three  thousand 
eighty-seconds. 

Eead  and  analyze  the  following  fractions : 

^^'   iV  J  T^'y  A?  M>  TT  ?  TlTJ   WV?  iM- 

13'  ftt;  T%^;  ^i^;  li^ir;  ^Hrr;  ^m. 

131.  Fractions  are  distinguished  as  Proper  and  Improper. 

A  Proper  Fraction  is  one  whose  numerator  is  less  than 
its  denominator ;  its  value  is  less  than  the  unit,  1.  Thus, 
tV>  a?  A>  Jf  ^^6  proper  fractions. 

An  Improper  Fraction  is  one  whose  numerator  equals 
or  exceeds  its  denominator ;  its  value  is  never  less  than  tlio 
unit,  1 .    Thus,  ^,  f,  Jjt,  •^-,  \^,  i^  are  improper  fractions 

123.  A  Mixed  Number  is  a  number  expressed  by  an  in- 
teger and  a  fraction;  thus,  4J-,  17J|,  9^  are  mixed  numbers* 

123.  Since  fractions  indicate  division,  all  changes  in  the 
terms  of  a  fraction  will  affect  the  value  of  that  fraction  ac- 
cording to  the  laws  of  division  ;  and  we  have  only  to  modify 
the  language  of  the  General  Principles  of  Division  (87)  by 
substituting  the  words  numerator,  denominator,  and  fraction. 


What  is  a  proper  fraction  ?    An  improper  fraction  ?    A  mixed  num' 
ber  ?    Wliat  do  fractions  indicate  ? 


REDUCTION.  89 

or  value  of  the  fraction,  for  the  words  dividend,  divisor,  and 
quotient,  respectively,  and  we  shall  have  the  following 

Gekeeal  Principles  of  Fractions. 

124*  Prin.  I.  Multiplying  the  numerator  multiplies  the 
fraction,  and  dividing  the  numerator  divides  the  fraction, 

Prin.  II.  Multiplying  the  denominator  divides  the  frao- 
fion,  and  dividing  the  denominator  multiplies  the  fraction, 

Prin.  III.  Multiplying  or  dividing  both  terms  of  the  frao* 
Hon  hy  the  same  number  does  not  alter  the  value  of  the  fraction. 

These  three  principles  may  be  embraced  in  one 

General  Law. 

125.  A  change  in  the  numerator  produces  a  like  change 
in  the  value  of  the  fraction  ;  but  a  change  in  the  denomina- 
tor produces  an  opposite  change  in  the  value  of  the  fraction^ 

REDUCTION. 

Case  L 

126.  To  reduce  firactions  to  their  lowest  terms. 

A  fraction  is  in  its  lowest  terms  when  its  numerator  and 
denominator  are  prime  to  each  other ;  that  is,  when  both 
terms  have  no  common  divisor. 

1.  Reduce  the  fraction  f^  to  its  lowest  terms. 

FiKST  OPERATION.  ANALYSIS.    Dividing  both  terms  of 

I     j^=:-|^=^=^,  ^^^5.  a  fraction  by  the  same  number  does 

not  alter  the  value  of  the  fraction  or 
qnotient,  (124,  III ;)  hence,  we  divide  both  terms  of  f  f ,  by  2,  both 
terms  of  the  result,  f-f,  by  2,  and  both  tei-ms  of  this  result  by  3.  As 
the  terms  of  f  are  prime  to  each  other,  the  lowest  terms  of  ||  are  |. 
We  have,  in  effect,  canceled  all  the  factors  common  to  the  numerator 
and  denominator. 

First  general  principle?  Second?  Third?  General  law?  What 
is  meant  by  reduction  of  fractions  ?  Case  I  is  what  ?  What  is  meant 
hj  lowest  terms  f    Give  analysis. 


90  FEACTIONS. 

SECOND  OPERATION.  In  this  operation  we  have  divided 

12  )  |-§-=i^,  Ans.  hoth  terms  of    the  fraction  by  their 

greatest    common    divisor,   (97,)  and 
thus  performed  the  reduction  at  a  single  division. 

Rule.   Cancel  or  reject  all  factors  common  to  both  numera- 
tor and  denommator.    Or, 
Divide  hath  terms  hy  their  greatest  common  divisor. 


Examples  fob  Peactice. 

2. 

Reduce 

Jll^    to  its  lowest  terms. 

Ans.  \, 

3. 

Reduce 

Iff    to  its  lowest  terms. 

Ans.  ^. 

4. 

Reduce 

j^    to  its  lowest  terms. 

Ans,  If 

6. 

Reduce 

f^    to  its  lowest  terms. 

6.  Reduce  ^iS%  ^^  its  lowest  terms. 

7.  Reduce  -j^^  to  its  lowest  terms. 

8.  Reduce  -^-^  to  its  lowest  terms. 

9.  Reduce  -f^ff  to  its  lowest  terms.  Ans.  JJ. 

10.  Reduce  Iffl  to  its  lowest  terms.  Ans.  fg- . 

11.  Reduce  ^^%  to  its  lowest  terms.  Ans.  H^. 

12.  Express  in  its  simplest  form  the  quotient  of  441  di* 
Tided  by  462.  Ans.  f}. 

13.  Express  in  its  simplest  form  the  quotient  of  189  di- 
vided by  273.  Ans.  A- 

14.  Express  in  its  simplest  form  the  quotient  of  1344  di- 
vided by  1536.  Ans.  }. 

Case  IL 
127.  To  reduce  an  improper  fraction  to  a  whole 

CI  mixed  number. 
1.  Reduce  ^^  to  a  whole  or  mixed  number. 

OPERATION.  Analysts.    Since 

5JLt  =  324  -^  15  =  2lT«r  =  211,  Ans.      15  fifteenths  equal 

1,  324  fifteenths  are 
equal  to  as  many  times  1  as  15  is  contained  limes  in  824,  which  is 
Slf'^y  times.     Or,  since  the  numerator  is  a  dividend  and  the  denom* 

Bule.    Case  II  is  what?    Give  explanation. 


REDUCTION".  91 

inator  a  divisor  ( 118 ),  we  reduce  the  fraction  to  an  equivalent  whole 
or  mixed  number,  by  dividing  the  numerator,  324,  by  the  denom- 
inator, 15. 

KuLE.  Divide  the  numerator  hy  the  denominator, 

1.  When  the  denominator  is  an  exact  divisor  of  the  numerator,  the  result  wiD  be 
A  whole  nu.nber. 

2.  In  all  answers  containing  fractions  reduce  the  ftactions  to  their  lowest  tenna 

Examples  foe  Peactice. 

2.  In  J^-  of  a  week,  how  many  weeks  ?  Ans.  1^. 

3.  In  4^  of  ^  bushel,  how  many  bushels  ?  Ans,  23f . 

4.  In  ^^  of  a  dollar,  how  many  dollars? 

5.  In  ^f-  of  a  pound,  how  many  pounds  ?  Ans,  54^. 

6.  Reduce  J-|4^  to  a  mixed  number. 

7.  Reduce  ^f-  to  a  whole  number. 

8.  Change  ^|p  to  a  mixed  number.  Ans.  18f . 

9.  Change  ^^  to  a  mixed  number. 

10.  Change  -^W^  to  a  mixed  number.     Ans.  1053f|. 

11.  Change  "^l\l^^  to  a  whole  number.       Ans.  7032. 

Case  III. 

128.  To  reduce  a  whole  mimber  to  a  fraction 
having  a  given  denominator. 

1.  Reduce  46  yards  to  fourths. 

OPERATION.  Akalysis-     Since  in  1  yard  there  are  4  fourths, 

46  in  46  yards  there  are  46  times  4  fourths,  which  are 

^  184  fourths  =  ^^.    In  practice  we  multiply  46, 

the  number  of  yards,  by  4,  the  given  denominator, 

•1^,  Ans,  and  taking  the  product,  184,  for  the  numerator  of 

a  fraction,  and  the  given  denominator,  4,  for  the 
denominator,  we  have  ^f^. 

Rule.  Multiply  the  tvhole  numher  ly  the  given  denomina- 
tor ;  take  the  product  for  a  numerator,  under  which  write 
the  given  denominator. 

Rule.    Case  III  is  what  ?    Give  explanation.    Rule. 


92  FRACTIONS. 

A  whole  number  is  reduced  to  a  fractional  form  by  writing  1  under  it  for  a  de- 
nominator ;  thus,  9  =  ^. 

Examples  for  Practice. 

2.  Reduce  25  bushels  to  eighths  of  a  bushel.     Ans.  -8-g-fi-. 

3.  Eeduce  63  gallons  to  fourths  of  a  gallon.      A?is,  ^^. 

4.  Reduce  140  pounds  to  sixteenths  of  a  pound. 

5.  In  56  dollars,  how  many  tenths  of  a  dollar  ?  Ans,  ^^, 

6.  Reduce  94  to  a  fraction  whose  denominator  is  9. 

7.  Reduce  180  to  seventy-fifths. 

8.  Change  42  to  the  form  of  a  fraction.  Ans.  ^, 

9.  Change  247  to  the  form  of  a  fraction. 

10.  Change  347  to  a  fraction  whose  denominator  shall 
be  14.  A71S.  -^-fl-S-. 

Case  IY. 

129.  To  reduce  a  mixed  number  to  an  improper 
fraction. 

1.  In  5|  dollars,  how  many  eighths  of  a  dollar  ? 

OPEEATION. 

54  Analysis.     Since  in  1  dollar  there  are  8  eighths, 

Q  in  5  dollars  there  are  5  times  8  eighths,  or  40 

—  eighths,  and  40  eighths  +  3  eighths  =  43  eighths, 

4g3.,  A7IS,  or  ^K 

Rule.  Multiply  the  whole  number  hy  the  denominator  of 
the  fraction;  to  the  product  add  the  numerator^  and  under 
the  sum  write  the  denominator. 

Examples  for  Practice. 

2.  In  4 J  dollars,  how  many  half  dollars  ?  Ans.  |. 
8.  In  71f  weeks,  how  many  sevenths  of  a  week  ? 

4.  In  341  j-  acres,  how  many  fourths  ?  Ans.  -LSj^-^. 

5.  Change  1^^  years  to  twelfths. 

6.  Change  56^^  to  an  improper  fraction.  Ans.  -V^. 

7.  Reduce  21^  to  an  improper  fraction.  Ans.  -^-fF-- 

8.  Reduce  225^  to  an  improper  fraction.  A71S.  ^l\^* 
* " 

Case  IV  is  what  ?    Give  explanation.    Rule. 


REDUCTION-.  93 

9.  In  96^<V,  how  many  one  hundred  twentieths  ? 

10.  In  1297^,  how  many  eighty-fourths ?    Ans.  ^^H^K 

11.  What  improper  fraction  will  express  400-|4  ? 

Case  V. 
130.  To  reduce  a  fraction  to  a  given  denominator. 

As  fractions  may  be  reduced  to  lower  terms  by  division, 
they  may  also  be  reduced  to  higher  terms  by  multiplication  ; 
and  all  higher  terms  must  be  multiples  of  the  lowest  terms. 
(103.) 

1.  Eeduce  }  to  a  fraction  whose  denominator  is  20. 

OPERATION.  Analysis.    First  divide  20,  the  required 

20  -f-  4  =  5  denominator,  by  4,  the  denominator  of  the 

o  ^   K  given  fraction,  to  ascertain  if  it  be  a  mul- 

—         ^^i^j  -4w5.         tiple  of  this  term,  4.     The  division  shows 

4x5  that  it  is  a  multiple,  and  that  5  is  the  factor 

which  must  be  employed  to  produce  this 

multiple  of  4.    We  therefore  multiply  both  terms  of  f  by  5,  ( 124,) 

and  obtain  ^f ,  the  desired  result. 

Rule.  Divide  the  required  denominator  hy  the  denomina- 
tor of  the  given  fraction,  and  multiply  both  terms  of  the 
fraction  by  the  quotient. 

Examples  for  Practice. 

2.  Reduce  f  to  a  fraction  whose  denominator  is  15. 

Ans.  ^j, 

5.  Reduce  -^  to  a  fraction  whose  denominator  is  35. 
4.  Reduce  -j^  to  a  fraction  whose  denominator  is  51. 

A71S.  If. 

6.  Reduce  %^  to  a  fraction  whose  denominator  is  150. 

6.  Reduce  J-|f  to  a  fraction  whose  denominator  is  3488. 

Ans,im- 

7.  Reduce  -^  to  a  fraction  whose  denominator  is  1000. 

Case  V  is  what?  How  are  fractions  reduced  to  higher  terms? 
What  are  all  higher  terms  ?     Give  analysis.     Rule. 


94  FEACTI0N8. 

Case  VL 

131.  To  reduce  two  or  more  fractions  to  a  com- 
mon denominator. 

A  Common  Denominator  is  a  denominator  common 
to  two  or  more  fractions. 

1.  Reduce  f  and  |  to  a  common  denominator. 

OPERATION.  Analysis.    Multiply  the  terms  of  the  first 

3x5  fraction  by  the  denominator  of  the  second,  and 

~r       K  ^^  2T  the  terms  of  the  second  fraction  by  the  denom- 

^  ^  ^  inator  of  the  first,  (124.)    This  must  reduce 

.2x4  esich.  fraction  to  the  same  denominator,  for 

^  zo  each  new  denominator  will  be  the  product  of 

^  '^  *  the  given  denominators. 

Rule.  Multiply  the  terms  of  each  fraction  by  the  denom- 
inators of  all  the  other  fractions. 

Mixed  numbers  must  first  be  reduced  to  Improper  fractions. 

Examples  for  Practice. 

2.  Reduce  |,  i,  and  |  to  a  common  denominator.   • 

^ns,  if,  ih  a 

3.  Reduce  ^  and  f  to  a  common  denominator. 

Ans,  f i,  ff, 
4  Reduce  ^,  ^,  and  f  to  a  common  denominator. 

M.ns.  |g5,  skof  ■fbO' 

5.  Reduce  -|,  |,  f ,  and  |-  to  a  common  denominator. 

A71S.    -J^,    8T8>    836"^    SSff 

6.  Reduce  ^,  ^,  and  f  to  a  common  denominator. 

^»«-  m,  m-  A 

7.  Reduce  |^,  2 J,  J,  and  -J^  to  a  common  denominator. 

Ans.  m,  m,  Hh  ^ 

8.  Reduce  IJ,  -f^,  and  4  to  a  common  denominator. 
Ans.  Jtf ,  U,  W 

Case  VI  is  what  ?    What  is  a  common  denominator  ?    Give  analysis. 
Rule. 


REDUCTIOI?-.  96 

Case  VII. 
132.  To  reduce  fractions  to  the  least  common  de- 
nominator. 

The  Least  Common  Denomiinator  of  two  or  more 
fractions  is  the  least  denominator  to  which  they  can  all  be 
reduced,  and  it  must  be  the  least  common  multiple  of  the 
lowest  denominators. 

1.  Eeduce  ^,  |,  and  -^  to  the  least  common  denominator. 

OPERATION.  Analysis.    First  find  the 

2    3       6  . .  8  . .  12  least  common  multiple  of  the 

2    2  4        2  given    denominators,    which. 

is   24.      This   must   be    the 

least  common  denominator  to 
which  the  fractions   can  be 
Ans,  reduced,  (III.)    Then  multi- 

ply the  terms  of  each  fraction 
by  such  a  number  as  will  re- 
duce the  fraction  to  the  denominator,  24.  Reducing  each  fraction  to 
this  denominator,  by  Case  V,  we  have  the  answer. 

Since  the  common  denominator  is  already  determined,  it 
is  only  necessary  to  multiply  the  numerators  by  the  multi- 
pliers. 

Rule.  I.  Find  the  least  common  multiple  of  the  given  de- 
nominators, for  the  least  common  denominator. 

II.  Divide  this  common  denominator  hy  each  of  the  given 
denominators,  and  multiply  each  numerator  hy  the  corres- 
.ponding  quotient.    The  products  loill  be  the  new  numerators. 

Examples  fok  Peactice. 

2.  Reduce  ^,  ^,  fj,  and  t^  to  their  least  common  de- 
nominator. Ans.  -^,  -^,  m,  j^. 

3.  Reduce  ^,  4?  A>  A  ^^  their  least  common  denomina- 
tor. Ans.  m  m^  *.  m- 

What  is  Case  VII  ?  What  must  be  the  least  common  denominator  1 
Give  analysis.     Rule,  first  step.     Second. 


96  FRACTIONS. 

4.  Beduce  f ,  ^,  J,  and  6  to  their  least  common  denomi- 
nator. Arts,  ^i^,  ^,  ill,  1^, 

5.  Reduce  5^,  2J,  and  If  to  their  least  common  denomi- 
nator. Arts,  ^y  ^,  ^. 

6.  Reduce  ^,  |,  4^,  and  J-  to  their  least  common  denomi- 
nator. Ans.  lii,  ffi,  III,  Hf. 

7.  Reduce  f ,  ^,  f,  2|-,  and  ^  to  their  least  common  de- 
nominator. Ans.  m,  -^y  -^,  m,  ^. 

8.  Change  f ,  3^,  3f ,  9,  and  -J  to  equivalent  fractions  hav- 
ing the  least  common  denominator. 

9.  Change  fi,  1^,  -J,  -J-J,  and  6  to  equivalent  fractions 
having  the  least  common  denominator. 

10.  Change  2yV»  |J,  4,  If,  ^,  and  |  to  equivalent  frac- 
tions having  the  least  common  denominator. 

11.  Reduce  f ,  |,  |-,  and  3^  to  a  common  denominator. 

12.  Reduce  -J,  \,  2f ,  and  |  to  a  common  denominator. 

13.  Reduce  {^,  -^y  f,  and  3-^  to  equivalent  fractions  hav- 
ing a  common  denominator.  A71S.  f|-,  -§-}-,  H,  |-J. 

14.  Change  ^,  |,  and  f  to  equivalent  fractions  having  a 
common  denominator.  Ans.  -f^^,  WW>  iWlr- 

15.  Change  -j^:?  7-|-,  f^,  and  5  to  equivalent  fractions  hav- 
ing a  common  denominator.  Ans.  ff,  ^^,  ff ,  -W. 

16.  Change  ^V»  ^i,  A,  7, 1,  and  IJ  to  equivalent  fractions 
having  a  common  denominator. 

ADDITION. 
133.  1.  What  is  the  sum  of  J,  |,  |,  and  |? 

OPEBATION.  Analysis.    Since  the 

1  4.  I  4_  I  4.  I  z=  Jgt  =  2,  ^W5.  S^^^""  fraxjtions  have  a 

common  denominator,  8, 
their  sum  may  be  found  by  adding  their  numerators,  1,3,  5,  and  7, 
and  placing  the  sum,  16,  over  the  common  denominator.  We  thus 
obtain  ^^^=2,  the  required  sum. 

2.  Add  ^,  A,  tV.  a.  and  ^.  Ans.  2^. 

3.  Add  ^,  rfj,  j\,  ^,  ^y,  and  {i.  Ans.  2 A- 

Give  first  explanation. 


ADDITION".  97 

4.  What  is  the  sum  of  ^,  A,  ^,  H.  if.  and  H? 
6.  What  is  the  sum  of  -^,  yV_^  ^^  ^^  and  |ff  ? 
6.  What  is  the  sum  of  ^,  ■^,  if^,  Hi,  and  f||  ? 

134.  1.  What  is  the  sum  of  |  and  |  ? 

OPEKATiON.  Analysis.    In 

|+|  =  ||  +  J^=  -^^iffl  =  H,  ^WS.  whole  numbers 

we  can  add  like 
numbers  only,  or  those  having  the  same  unit  value ;  so  in  fractions 
we  can  add  the  numerators  when  they  have  a  common  denominator, 
but  not  otherwise.  As  f  and  f  have  not  a  common  denominator,  first 
reduce  them  to  a  common  denominator,  and  then  add  the  numerators, 
27  +  10=37,  the  same  as  whole  numbers,  and  place  the  sum  over  the 
common  denominator. 

Rule.  I.  When  necessary,  reduce  the  fractions  to  their 
least  common  denominator. 

II.  Add  the  numerators,  and  place  the  sum  over  the  com- 
mon denominator. 

If  the  amount  be  an  improper  fraction,  reduce  it  to  a  whole  or  a  mixed  number. 

Examples  for  Practice. 

2.  Add  J  to  f .  Ans,  ||. 

3.  Add  i-  to  ii.  Ans.  1^, 

4.  Add  },  \,  f ,  and  ^.  Ans.  l-^. 

5.  Add  H,  fJ.  and  ^.  Ans.  1^^, 

6.  Add  ^0.  A,  A.  and  ^.  Ans.  f. 

7.  Add  ^,  m,  U,  h  and  |.  Ans.  3^. 

8.  Add  },  h  h  h  h  h  h  h  and  ^,  Ans.  7^^. 

9.  Add  7i,  5|,  and  lOf. 

OPERATION.  Akalysis,    The  sum  of  the  frac- 

•J^  4-  I  +     }  =     1^  tions  ^,  f ,  and  f  is  IH  ;  tlie  sum  of 

7  -i_  5  _|_  10  ^  22  *^®  integers,  7,  5,  and  10,  is  22  ; 

and  the  sum  of  both  fractions  and 

Ans.  23\i  integers  is  23^^.    Hence, 

Give  second  explanation  ?    Rule,  first  step.  "  Second. 
R.P.  5 


98  FEACTIONS. 

To  add  mixed  numbers,  add  the  fractions  and  integers 
separately,  and  tlien  add  their  sums. 

If  the  mixed  numbers  are  email,  they  may  be  reduced  to  improper  fractions,  and 
then  added  after  the  usual  method. 

10.  What  is  the  sum  of  14|,  3^^,  If,  and  IJ  ?  Ans.  21^. 

11.  What  is  the  sum  of  -J,  1^^,  10|,  and  5  ?      Ans.  IS^V 

12.  What  is  the  sum  of  17i,  18^,  and  26^? 

13.  What  is  the  sum  of  A^  tt^  ^i  3,  and  iJ  ? 

14.  What  is  the  sum  of  125^,  327^,  and  25i  ?  Ans.  478^. 

15.  What  is  the  sum  of  ^H,  |J,  1^,  «,  and  «|? 

Ans.  3ttJ. 

16.  What  is  the  sum  of  3^^,  2|-|,  40|,  and  10^  ? 

17.  Bought  3  pieces  of  cloth  containing  125-J,  96 f,  and 
48f  yards ;  how  many  yards  in  the  3  pieces  ? 

18.  If  it  take  5^-  yards  of  cloth  for  a  coat,  3^  yards  for  a 
pair  of  pantaloons,  and  J  of  a  yard  for  a  vest,  how  many 
yards  will  it  take  for  all  ?  Ans.  ^r}^. 

19.  A  farmer  divides  his  farm  into  5  fields  ;  the  first  con- 
tains 26y^  acres,  the  second  40^  acres,  the  third  514  acres, 
the  fourth  59f  acres,  and  the  fifth  62|-  acres ;  how  many 
acres  in  the  farm  ?  Ans.  241|^. 

20.  A  speculator  bought  175|  bushels  of  wheat  for  205 J 
dollars,  325^  bushels  of  barley  for  296}  dollars,  270^1  bush- 
els of  corn  for  200^  dollars,  and  43 7-^^  bushels  of  oats  for 
156f|-  dollars  ;  how  many  bushels  of  grain  did  he  buy,  and 
how  much  did  he  pay  for  the  whole  ?     j  ^«  j  1 2^^ A  bushels. 

•  (  859}S  dollai's. 

SUBTKACTION. 

135.  1.  From  ^  take  A- 

OPERATION.  Analysis.     Since  the  given 

^1^  —  ^  =  -j*jj  =  ^,  Ans.         fractions  have  a  common  denom- 

inator,  10,  we  find  the  difference 
by  subtracting  3,  the  less  numerator,  from  7,  the  greater,  and  write 


How  are  mixed  numbers  added  %    Give  note. 


SUBTRACTION".  99 

the  remainder,  4,  over  the  common  denominator,  10.    We  thus  obtain 
y%  =  f ,  the  required  difference. 


2.  From  |  take  |. 

^^5.  ^. 

3.  From  ^  take  ^. 

^/is.  1^. 

4.  From  fO-  take  ^. 

Ans.  \\. 

5.  From  f|  take  ff . 

Ans.  \^. 

6.  From  -^  take  ^i^. 

Ans,  J. 

7.  From  Ml  take  iif 

Ans.  ■^. 

136.  1.  From  f  take  f. 

OPERATION. 

Analysis. 

4         4  —  42           P-  —   32-30   - 

-   -A-   -Xr        Atl^ 

As  in  whole 

¥         t         T6           36               3  6 

-ir^  —  T«>  -a./*6. 

numbers,  we 

can  subtract  like  numbers  only,  or  those  having  the  same  unit  value, 
so,  we  cau  subtract  fractions  only  when  they  have  a  common  denom- 
inator. As  I  and  f  have  not  a  common  denominator,  we  first  reduce 
them  to  a  common  denominator,  and  then  subtract  the  less  numerator, 
30,  from  the  greater,  32,  and  write  the  difference,  2,  over  the  common 
denominator,  36. 

EuLE.  I.  Wlien  necessary y  reduce  the  fractions  to  a  com- 
mon denominator. 

11.  Subtract  the  numerator  of  the  subtrahend  from  the 
numerator  of  the  minuend,  and  place  the  difference  over  the 
co7nmon  denominator. 

Examples  for  Peactice. 

2.  From  \  take  |.  Ans.  -fy. 

3.  From  ^  take  I .  Ans.  j^. 

4.  Subtract  -^  from  f .  Ans.  ■^. 

5.  Subtract  -^  from  -^.  Ans.  ^. 

6.  Subtract  f|  from  ^^.  Ans.  -^. 

7.  Subtract  ^^^  from  f^.  Ans.  ^^. 

8.  What  is  the  diSereuce  between  9-J  and  2}  ? 

Give  explanations.     Rule,  first  step.     Second. 


100  FEACTIONS. 

OPERATION.  Analysis.    First  reduce  the  fractional 

91  z=  9  A-  parts,  -|-  and  f ,  to  a  common  denominator, 

03  __  2_^  ^^'    ^^^^®  ^®  cannot  take  -f^  from  j\,  we 

add  1  =  If  to  j\,  which  makes  f|,  and 

6^,  Ans.  3»^  from  if  leaves  y'^.     We  now  add  1  to 

the  2  in  the  subtrahend,  (50,)  and  say,  3 
from  9  leaves  6.    We  thus  obtain  6xV,  the  difference  required. 

Hence,  to  subtract  mixed  numbers,  we  may  reduce  the 
fractional  parts  to  a  common  denominator,  and  then  subtract 
the  fractional  and  integral  parts  separately.    Or, 

We  may  reduce  the  mixed  numhers  to  improper  fractions^ 
and  subtract  the  less  from  the  greater  by  the  usual  method, 

9.  From  8 J  take  S-J.  Ans.  4ff. 

10.  From  25|-  take  9^.  Ans,  16^^. 

11.  From  ^  take  \^, 

12.  Subtract  1^  from  6. 

13.  Subtract  120^  from  450^.  Ans,  330|f. 

14.  Subtract  ^  from  3^V  Ans,  3^. 

15.  Find  the  difference  between  49  and  75 J-. 

16.  Find  the  difference  between  327|  and  196|. 

17.  From  a  cask  of  wine  containing  31^  gallons,  17|-  gal- 
lons were  drawn  ;  how  many  gallons  remained  ?  Ans.  13 J. 

18.  A  farmer,  having  450^  acres  of  land,  sold  304| 
acres  ;  how  many  acres  had  he  left?  Ans.  145^^. 

1 9.  If  flour  be  bought  for  6J-  dollars  per  barrel,  and  sold 
for  7|  dollars,  what  will  be  the  gain  per  barrel  ? 

20.  From  the  sum  of  4  and  3^  take  the  difference  of  4-J 
and  5i.  Ans.  3|f . 

21.  A  man,  having  25|  dollars,  paid  Q\  dollars  for  coal, 
2^  dollars  for  dry  goods,  and  J  of  a  dollar  for  a  pound  of 
tea ;  how  much  had  he  left  ?  Ans.  Iiejj. 

22.  What  number  added  to  2|will  make  7J?  AnsA\^. 

23.  What  fraction  added  to  {^  will  make  ^  ?  Ans.  ^. 

In  how  many  ways  may  mixed  numbers  be  subtracted  ?  Wliat  arc 
they? 


MULTIPLICATION.  101 

24.  A  gentleman,  having  2000  dollars  to  divide  among 
his  three  sons,  gave  to  the  first  912J  dollars,  to  the  second 
545^  dollars,  and  to  the  third  the  remainder ;  what  did  the 
third  receive  ?  Ans.  $542^^. 

25.  Bought  a  quantity  of  coal  for  136^  dollars,  and  of 
lumber  for  350f  dollars.  I  sold  the  coal  for  184^  dollars, 
and  the  lumber  for  41 6f  dollars.   What  was  my  whole  gain  ? 

Ans.  $114^. 

MULTIPLICATION. 

Case  I. 
137.  To  miiltiply  a  fraction  by  an  integer. 

1.  If  1  yard  of  cloth  cost  f  of  a  dollar,  how  much  will  5 

yards  cost  ? 

OPERATION.  Analysis.     Since  1  yard  cost 

I  X  5  =  ^-  =-  ^i,  Ans.  3  fourths  of  a  dollar,  5  yards 

will  cost  5  times  3  fourtJis  of  a 
dollar,  or  15  fourths,  equal  to  3f  dollars.  A  fraction  is  multiplied  by 
multiplying  its  numerator,  (124.) 

2.  If  1  gallon  of  molasses  cost  ^  of  a  dollar,  how  much 
will  5  gallons  cost  ? 

OPERATION.  Analysis.    Since  5,  the  muU 

■^  X  5  =:  J  =  If ,  ^725.  tiplier,  is  a  factor  of  20,  the  de- 

nominator,  of  the  multiplicand, 
we  perform  the  multiplication  by  dividing  the  denominator,  20,  by  the 
multiplier,  5,  and  we  have  |,  equal  to  1^  dollars.  A  fraction  is  multi- 
plied by  dividing  its  denominator,  (124.)    Hence, 

MuUi^jlying  a  fraction  consists  in  multiplying  its  nu- 
merator, or  dividing  its  denominator. 

Always  divide  the  denominator  when  it  is  exactly  divisible  by  the  multiplier. 

Examples  for  Practice. 

3.  Multiply  f  by  5.  ,  Ans.^  =  ^. 

4.  Multiply  yV  by  7.  Ans.  l|i. 

Case  I  is  what  ?    Give  explanations.    Deduction. 


102 


FRACTIONS. 


5.  Multiply  T^  by  12. 

6.  Multiply  ^  by  63. 

7.  Multiply  5}  by  9. 

Ans.  74. 
Ans,  15. 

OPEKATION. 

9                Or, 
45        ^x9=^=49i. 

Analysis.     In    multiply, 
ing  a  mixed  number,  first  mul- 
tiply the  fractional  part,  and 
then  the  integer,  and  add  the 
two  products;  or,  reduce  the 
mixed  number  to  an  improper 
fraction,  and  then  multiply  it. 

8.  Multiply  7|  by  12. 

9.  Multiply  ^  by  8. 

10.  Multiply  yl^  by  51. 

11.  Multiply  15|  by  16. 

12.  Multiply  m  by  22. 

Ans.  91|. 
Ans.  6-^. 
Ans.        2. 
Afis.    250. 
Ans.    16|-. 

13.  If  a  man  earn  8^<j  dollars  a  week,  how  many  dollars 
will  he  earn  in  12  weeks  ? 

14.  What  will  9  yards  of  silk  cost  at  ^  of  a  dollar  per 
yard? 

15.  What  will  27  bushels  of  barley  cost  at  |  of  a  dollar 
per  bushel  ?  Ans.  23|  dollars. 


Case  II. 

138.  To  multiply  an  integer  by  a  fraction. 

1.  At  75  dollars  an  acre,  how  much  will  |  of  an  acre  of 
land  cost  ? 


PIKST  OPERATION. 
5  )  75    price  of  an  acre. 

1 5    cost  of  ^  of  an  acre. 

3 


Ans,  45    cost  of  f  of  an  acre. 


Analysis.  8  fifths  of  an  acre 
will  cost  three  times  as  much  as 
1  fifth  of  an  acre.  Dividing  75 
dollars  by  5,  we  have  15  dollars, 
the  cost  of  I  of  an  acre,  which 
we  multiply  by  3,  and  obtain  45 
dollars,  the  cost  of  f  of  an  acre. 


Explain  the  process  of  multiplying  mixed  numbers.    What  is  Case 
II  ^    Give  first  explanation. 


MULTIPLICATION. 


103 


6BC0KD  OPERATION. 
75  price  of  1  acre. 

3 


5  )  225  cost  of  3  acres. 
Ans.     45    "    "      1^  of  an  acre. 


Or,  multiplying  tlie  price 
of  1  acre  by  3,  we  have  the 
cost  of  3  acres  ;  and  as  ^  of 
3  acres  is  the  same  as  |  of 
1  acre,  we  divide  the  cost 
of  3  acres  by  5,  and  we  have 
the  cost  of  f  of  an  acre,  the 
same  as  in  the  first  opera- 
tion.   Hence, 

Multiplying  by  a  fraction  co7isists  in  multiplying  ly  the 
numerator  and  dividing  ly  the  denominator  of  the  multiplier, 

15 

J^^  By  using  the  vertical  line  and  cancellation,  we 

3 


$ 


Bhall  shorten,  and  combine  both  operations  in 
one. 


45,  Ans, 


Examples  for  Practice. 


2.  Multiply  3  by  f . 

3.  Multiply  100  by  A- 

4.  Multiply  105  by  J|. 

5.  Multiply  19  by  i|. 

6.  Multiply  24  by  6|, 

OPEKA.TION. 
24 

15  =  f  of  24  • 

^44 


Or, 


150,  Ans, 


53 


159,  Ans, 


7.  Multiply  42  by  9}. 

8.  Multiply  80  by  14^. 

9.  Multiply  156  by  f|. 
10.  At  8  dollars  a  bushel,  what  will  f  of  a  bushel  of  clover 

seed  cost  ? 


Ans,  IJ. 
Ans.  64f. 
Ans.  85. 
Ans,  5if. 

Analysis.  Mul- 
tiply by  the  integer 
and  fraction  sepa- 
rately, and  add  the 
products ;  or,  reduce 
the  mixed  number 
to  an  improper  frac- 
tion, and  then  mul 
tiply  by  it. 

Ans,  409f 
Ans,  1165. 
Ans.    108. 


Give  second  explanation.    Note.     Deduction. 


104  FRACTIONS. 

11.  If  a  man  travel  36  miles  a  day,  how  many  miles  will 
he  travel  in  lOf  days  :  Ans.  384  miles. 

12.  If  a  village  lot  be  worth  450  dollars,  what  is  -^  of  it 
worth?  Ans,  262^  dollars. 

13.  At  16  dollars  a  ton,  what  is  the  cost  of  2}  tons  of  hay  ? 

Case  III. 
139.  To  multiply  a  fraction  by  a  fraction, 
1.  At  f  of  a  dollar  per  bushel,  how  much  will  J  of  a  bushe\ 


of  com  cost 

9 

OPEEATiON.                                          Analysis. 

1st  step,          1  -^  4  =  y^,  cost  of  }  of  a  busheL             Since  1  bush. 

SdBtep,          yV  X  3  =  ^%    ''  ''  1  ''  ''   "                  ^1  ^««*  t  of  a 

Whole  work,  1    X  f  =  A  =  i,  AnS.                            ^'^^l\  ^  ^^.^^ 
^         *        *•*        *                                     bushel     will 

Or,      $ 

%                           cost  f  times  f  of  a  dollar,  or  3  times 

h 

^                             ^  of  f  of  a  dollar.     Dividing  |  of  a 

^                                     doll  AT  by  Ai  WP!  Viavft  -jT^^  fhft  rnst.  nf  X 

2 

1  =  J,  ^7^5.       of  a  bushel.    A  fraction  is  divided  by 

multiplying  its  denominator,  (124.) 
Multiplying  the  cost  of  \  of  a  bushel  by  3,  we  have  -^^  of  a  dollar,  the 
cost  of  I  of  a  bushel.  It  will  readily  be  seen  that  we  have  multiplied 
together  the  two  numerators,  2  and  3,  for  a  new  numerator,  and  the 
two  denominators,  3  and  4,  for  a  new  denominator,  as  shown  in  the 
whole  work  of  the  operation.  Hence,  for  multiplication  of  fractions, 
we  have  this  general 

EuLE.  I.  Reduce  all  integers  and  mixed  numlers  to  im* 
proper  fractions, 

II.  Multiply  together  the  numerators  for  a  neio  numerator 
and  the  denominators  for  a  new  denominator. 

Cancel  all  foctors  common  to  numerators  and  denominators. 

2.  Multiply  I  by  f  Ans.    J. 

a  Multiply  J  by  |.  Ans.  ^, 

4.  Multiply  tt  by  -11.  Ans,  ■^. 

5.  Multiply  4  by  \.  Ans.  3|. 

What  is  Case  III  ?  Give  explanation.  Rule,  first  step  ?  Second  ! 
What  shall  be  done  with  common  factors  ? 


MULTIPLICATION?^. 


105 


»^ 

$ 

*<r0 

7 

6 

$ 

$ 

t 

$ 

t 

t 

4: 

$ 

$ 

30 

7  =  ^. 

6.  What  is  the  product  of  ^,  f,  |,  and  i  ?    ^?Z5.     ■^. 

7.  What  is  the  product  of  1|,  f,  2,  and  5-}?    J.?is.  11 J^. 

8.  What  is  the  product  of  f  of  ^,  |  of  |  of  |,  and  \  of 
If? 

OPERATION.  Or, 

—  X  — X  — X  — X  — X  — X— =— ,  ^ns, 
4     10^  Q      $      $      t      0     30 

Fractions  with  the  word  of  between  them  are  sometimes 
called  compound  fractions.  The  word  of  is  simply  an  equiv- 
alent for  the  "sign  of  multiplication,  x,  and  signifies  that 
th3  numbers  between  which  it  is  placed  are  to  be  multiplied 
together. 

9.  Multiply  ^  of  2i  by  \  of  7^.  Ans.  lf|. 

10.  Multiply  f  of  16  by  ^  of  26f.  Ans,  85^. 

11.  What  is  the  product  of  3,  J  of  -f-,  and  \  of  3J  ? 

12.  What  is  the  value  of  2 J  times  f  of  ^  of  1^  ?     Ans,  2. 

13.  What  is  the  value  of  J  of  ^  of  If  times  |  of  8  ? 

14.  What  is  the  product  of  12  J  multiplied  by  5^  times  6f  ? 

Ans,  464^. 

15.  At  f  of  a  dollar  per  yard,  what  will  |  of  a  yard  of 
cloth  cost  ?  Ans.  ^oi2i,  dollar. 

16.  If  a  man  own  f  of  a  vessel,  and  sell  f  of  his  share, 
what  part  of  the  whole  vessel  will  he  sell  ? 

17.  When  oats  are  worth  |-  of  a  dollar  per  bushel,  what  is 
}  of  a  bushel  worth  ? 

18.  What  will  7}  pounds  of  tea  cost,  at  |  of  a  dollar  per 
pound  ?  Ans.  4J  J  doUars. 

19.  What  is  the  product  of  9^  by  4|  ? 


39f  product  by  4. 


Or,  9f  x4|  =  —  X— =  46. 


Ans.  46 


"4|. 


What  doea  **  of"  signify  when  placed  between  two  fractions  ?    What 
a  compound  fraction  t 

6* 


106  FRACTIONS. 

To  multiply  mixed  numbers  together,  either  multiply  by 
the  integer  and  fractional  part  separately,  and  then  add 
their  products  ;  or,  reduce  both  numbers  to  improper  frac- 
tions, and  then  multiply  as  in  the  foregoing  rule. 

20.  Multiply  12f  by  8|.  A7is.  108|. 

21.  What  cost  6|-  cords  of  wood,  at  2f  dollars  a  cord  ? 

22.  What  cost  f  of  2J  tons  of  hay,  at  11^  dollars  a  ton  ! 

Ans.  $21^^. 

23.  What  will  8|  cords  of  wood  cost,  at  2f  dollars  per 
cord?  Ans.  22 JJ  dollars. 

24.  What  must  be  paid  for  |  of  6|^  tons  of  coal,  at  |  of  7} 
dollars  per  ton  ? 

25.  A  man  owning  J  of  a  farm,  sold  ^  of  his  share ;  what 
part  of  the  whole  farm  had  he  left  ?  Ans.  ^. 

26.  Bought  a  horse  for  125}  dollars,  and  sold  him  for  ^  of 
what  he  cost ;  what  was  the  loss  ?  Ans.  $25^. 

27.  A  owned  f  of  123|-  acres  of  land,  and  sold  |  of  his 
share  ;  how  many  acres  did  he  sell.  Ans.  49^. 

28.  If  a  family  consume  IJ-  barrels  of  flour  a  month,  how 
many  barrels  will  five  such  famiHes  consume  in  4^  months? 

DIVISION. 

Case  I. 
140.   To  divide  a  firaction  by  an  integer. 

1.  If  my  horse  eat  ^  of  a  ton  of  hay  in  3  months,  what 
part  of  a  ton  will  last  him  1  month  ? 

OPERATION.  Analysis.    If  he  eat  i%  ot  a  ton  in 

_t^  _i_  3  =  rAr,  Ans.  S  months,  in  1  month  he  will  eat  ^  of 

^  of  a  ton,  or  -^^  divided  by  3.     Since 

a  fraction  is  divided,  by  dividing  its  numerator,  (124,)  divide  the 

numerator  of  the  fraction,  ^jj,  by  8,  and  have  y^^. 

2.  If  3  yards  of  ribbon  cost  -J-  of  a  dollar,  what  will  1  yard 

cost  ? 


Case  I  is  what  ?    Give  first  explanation. 


DIVISION.  lOT 

OPERATION.  Analysis.    Here  we  cannot  exactly 

t  -^  3  =:  Jg^  Ans.  divide  tlie  numerator  by  3  ;  but,  since  a 

fraction  is  divided  by  multiplying  the 

denominator,  (124,)  we  multiply  the  denominator  of  the  fraction,  f , 

by  3,  and  have  ^q,  the  required  result.     Hence, 

Dividing  a  fraction  consists  in  dividing  its  numerator,  or 
multiplying  its  denominator. 

We  divide  the  numerator  when  it  is  exactly  divisible  by  the  divisor ;  otherwise 
w3  multiply  the  denominator. 

Examples  foe  Pkactice. 

3.  Divide    f    by    2.  Ans.    f 

4.  Divide  ^^  by    3.  Ans,     \, 

5.  Divide  f|  by    5.  Ans.   ^. 

6.  Divide  ^  by  25. 

7.  Divide  ^  by  14.  A?is.  ^. 

8.  Divide  f^  by  21.  A7is.  ^. 

9.  If  6  pounds  of  sugar  cost  |  of  a  dollar,  what  will  1 
pound  cost  ? 

10.  At  7  dollars  a  barrel,  what  part  of  a  barrel  of  flour 
can  be  bought  for  -J  of  a  dollar  ?  Ans.  ^. 

11.  If  a  yard  of  cloth  cost  5  dollars,  what  part  of  a  yard 
can  be  bought  for  -f  of  a  dollar?  Ans.  -^. 

12.  If  9  bushels  of  barley  cost  7-^  dollars,  what  will  1 
bushel  cost  ? 

OPERATION. 

-,     „  g  We  reduce  the  mixed  number  to  an  improper 

'6  ^~~    &  fraction,  and  divide  as  before. 

^^9=i,Ans. 

13.  If  12  barrels  of  flour  cost  76|-  dollars,  what  will  1 
barrel  cost  ? 

OPERATION.  Analysis.     We  first  divide  as  in  simple 

12  )  764  numbers,  and  have   a  remainder  of  4|. 

77      .  Reduce  this   remainder  to   an    improper 

6f,  Ans.  fraction,  -«/,  which  divide  (as  in  Ex.  1), 

and  annex  the  result,  f ,  to  the  partial  quotient,  6,  and  we  have  6|,  the 

required  result. 

Give  second  explanation.    Deduction. 


108  FEACTIONS. 

14.  How  many  times  will  16 J  gallons  of  cider  fill  a  vessel 
that  holds  3  gallons  ?  Ans.  5^. 

15.  If  9  men  consume  f  of  9|  pounds  of  meat  in  a  day, 
how  much  does  each  man  consume  ?    Ans.  ^  ot  s.  pound. 

16.  A  man  paid  I99||  for  4  cows ;  how  much  was  that 
apiece  ?  Ans,  $24f|. 

Case  IL 

141.  To  divide  an  integer  by  a  fraction. 

1.  At  f  of  a  dollar  a  yard,  how  many  yards  of  cloth  can 
be  bought  for  12  dollars  ? 

FIRST  OPERATiOsr.  ANALYSIS.     As  many  yards  as  |  of  8 

12  dollar,  tlie  price  of  1  yard,  is  contained 

^  times  in  13  dollars.     Integers  cannot  be 

divided  by  fourt/is,  because  they  are  not 

*^  /  ^"  of  the  same  denomination.     Reducing  12 

16  yards.  dollars  to  fourths  by  multiplying,  we  have 

48  fourths;  and  3  fourths  is  contained  in 
^fourths  16  times,  the  required  number  of  yards. 

SECOND  OPERATION.  ANALYSIS.     We  divide  the  integer  by 

3  )  12  the  numerator  of  the  fraction,  and  multi- 

T  ply  the  quotient  by  the  denominator,  vv^hich 

produces  the  same  result  as  in  the  first 

^  operation.     Hence, 

16  yards. 

Dividing  ly  a  fraction  consists  in  multiplying  by  the  de^ 
nominator y  and  dividing  hy  the  numerator  of  the  divisor. 

Examples  for  Practice. 

2.  Divide    18  by  |.  Ans,    48. 

3.  Divide    63  by  -^^  Ans,  117. 

4.  Divide    42  by  ^.  Ans,    49. 

5.  Divide  120  by  ■^,  Ans.  205f 

6.  Divide  316  by  ^g.  Ans.  877f 

Case  II  is  what  ?    Give  first  explanation.    Second.    Deduction. 


DIVISION.  109 

7.  How  many  bushels  of  oats,  worth  f  of  a  dollar  per 
bushel,  will  pay  for  f  of  a  barrel  of  flour,  worth  9  dollars  a 
barrel  ?  Ans.  15. 

8.  If  f  of  an  acre  of  land  sell  for  21  dollars,  what  will  an 
acre  sell  for  at  the  same  rate  ?  Atis.  $49. 

9.  When  potatoes  are  worth  |-  of  a  dollar  a  bushel,  and 
corn  I  of  a  dollar  a  bushel,  how  many  bushels  of  potatoes 
are  equal  in  value  to  16  bushels  of  corn?  Ans.  22-J-. 

10.  If  a  man  can  chop  2f  cords  of  wood  in  a  day,  in  how 
many  days  can  he  chop  22  cords  ? 

OPERATIOK. 

3f  =  ¥ 
22  Analysis.    "We  reduce  the  mixed  number 

.  to  an  improper  fraction,  and  then  divide  the 

integer  in  the  same  manner  as  by  a  proper 

11 )  88  fraction. 

Ans.  8    days 

11.  Divide  75  by  13f  Ans.  5f| . 

12.  Divide  149  by  2^.  Ans.  6^. 

13.  A  farmer  distributed  15  bushels  of  corn  among  some 
poor  persons,  giving  them  If  bushels  apiece ;  among  how 
many  persons  did  he  divide  it  ? 

14.  Divide  |  of  320  by  |  of  9f  Ans.  25f . 

15.  Bought  I  of  7^  cords  of  wood  for  J  of  132  ;  how  much 
did  1  cord  cost?  Ans.  $3-J-. 

16.  A  father  divided  183  acres  of  land  equally  among  his 
sons,  giving  them  45f  acres  apiece ;  how  many  sons  had  he  ? 

Ans.  4. 

Case  III. 
142.  To  divide  a  fraction  by  a  fraction. 

1.  How  many  pounds  of  tea  can  be  bought  for  -}-J  of  a 
dollar,  at  I  of  a  dollar  a  pound  ? 

How  divide  by  a  mixed  number  ?    Case  Til  is  what  ? 


110  FEACTIONS. 

OPERATION.  ^  Analysis.     As 

First  step,  ^x3=f-|  many  pounds  as  f 

SecondBtep,  ||_^2  =  |}=:lf.  of  a  dollar  is  con- 

11      2      11      3      11  tained  times  in  f| 

Wholework,  — -^-=— X^  =  -^-  =  l|,  Ans.      of  a  dollar.      1  is 

14      6      l^^   4,       b  contained  in  ii,  H 

times,  and  i  is  con- 
tained in  fl  3  times  as  many  times  as  1,  or  3  times  }|,  wliicli  is  f  | 
limes,  wliicli  is  the  number  of  pounds  that  could  be  bought  at  |^  of  a 
dollar  per  pound ;  but  f  is  contained  but  ^  as  many  times  as  J,  and 
fl  divided  b/  3  gives  ff,  equal  to  if  times,  or  the  number  of  pounds 
that  can  be  bought  at  |  of  a  dollar  per  pound. 

We  see  in  the  operation  that  we  have  multiplied  the  dividend  by 
the  denominator  of  the  divisor,  and  divided  the  result  by  the  numer- 
ator or  the  divisor,  which  is  in  accordance  with  140  for  dividing  a 
fraction.  Hence,  by  inverting  the  terms  of  the  divisor,  the  two  frac- 
tions will  stand  in  such  relation  to  each  other  that  we  can  multiply 
together  the  two  upper  numbers  for  the  numerator  of  the  quotient, 
and  the  two  lower  numbers  for  the  denominator,  as  shown  in  the 
operation.     For  division  of  fractions,  we  have  this  general 

Rule.  I.  Reduce  integers  and  mixed  numhers  to  improper 
fractions. 

II.  Invert  the  terms  of  tlie  divisor,  and  proceed  as  in  mul- 
tiplication. 

1.  The  dividend  and  divisor  may  be  reduced  to  a  common  denominator,  and  the 
numerator  of  the  dividend  be  divided  by  the  numerator  of  the  divisor ;  this  will  giv« 
the  same  result  as  the  rule. 

2.  Apply  cancellatiou  w^here  practicable. 

Examples  foe  Peactice. 

2.  Divide  J  by  J. 

3.  Divide  I  by  |. 
4  Divide  {  by  ■^. 

5.  Divide  ^  by  -^. 

6.  Divide  f  by  fj. 

7.  How  many  times  is  -f-  contained  in  f? 

8.  How  many  times  is  4  contained  in  1-|  ? 

llule,  first  step.    Second.    Wliat  other  method  is  mentioned 


Ans. 

If 

Ans. 

H- 

Ans. 

«• 

Ans. 

W- 

Ans. 

«• 

Ans. 

i,V 

Ahs. 

3t. 

DIVISION.  Ill 

9.  How  many  times  is  yV  contained  in  fj  ?     A71S.  2f . 

10.  How  many  times  is  ^  contained  in  J-f  ? 

11.  How  many  times  is  J  of  }  contained  in  f  of  2J? 

12.  What  is  the  quotient  of  ^  of  4,  divided  by  f  of  3 J  ? 

13.  What  is  the  quotient  of  ^  of  |  of  36  divided  by  1^ 
times  I  ?  A71S.  di. 

14.  What  is  the  value  of  -^ 


? 


OPERATION.  ,  This  example 

3|__^_7_^35_^       $l_         .  is  only  another 

4l~~'^~2  *   8~^^$0~~^^      ^^*  ^^^^    ^^^    ®^' 

5  pressing    divis- 

ion of  fractions ;  it  is  sometimes  called  a  complex  fraction,  and  the 
process  of  performing  the  division  is  called  reducing  a  complex  frac- 
tion to  a  simple  one. 

We  simply  reduce  the  upper  number  or  dividend  to  an  improper 
fraction,  and  the  lower  number,  or  divisor,  to  an  improper  fraction, 
and  then  divide  as  before. 


15.  What  is  the  value  of  ||  ?  Ans.  f  f. 

16.  What  is  the  value  of  ^?  Ans.  20. 

17.  What  is  the  value  of  g  ?  Ans.  ^. 

18.  What  is  the  value  of  ^^  ?  Ans.     1. 

19.  What  is  the  value  of  }  f  j;  ?  Ans.  i. 

I  of  4J^ 

20.  If  a  horse  eat  -|  of  a  bushel  of  oats  in  a  day,  in  how 
many  days  will  he  eat  5 J  bushels  ?  Ans.  14. 

21.  If  a  man  spend  If  dollars  per  month  for  tobacco,  in 
what  time  will  he  spend  lOf  dollars  ?       Ans.  6|-  months. 


What  is  a  complex  fraction  ? 


112  FBACTIOKS. 

22.  How  many  times  will  4f  gallons  of  camphene  fill  a 
vessel  that  holds  J  of  f  of  1  gallon  ?  Ans.  10|-. 

23.  If  14  acre?  of  meadow  land  produce  32f  tons  of  hay, 
how  many  tons  will  5  acres  produce?  A^is.  11|. 

24.  If  2  yards  of  silk  cost  $3 J,  how  much  less  than  $17 
will  9  yards  cost  ?  Ans.  $2f . 

25.  If  f  of  a  yard  of  cloth  cost  3^  of  a  dollar,  what  will 
1  yard  cost  ? 

26.  A  man,  having  $10,  gave  f  of  his  money  for  clover 
seed  at  13^  a  bushel ;  how  much  did  he  buy  ?    Ans.  2  bush. 

27.  How  many  tons  of  hay  can  be  purchased  for  IllQ^ig-, 
at  $9f  per  ton?  Ans.  12^. 

Promiscuous  Examples. 

1.  Eeduce  J,  f ,  f ,  and  J  to  equivalent  fractions  whose 
denominators  shall  be  24.  Ans.  Jf ,  f|,  ^,  ■^. 

2.  Change  4  to  an  equivalent  fraction  having  91  for  its 
denominator.  Ans.  ff* 

3.  rind  the  least  common  denominator  of  f ,  If,  ^  of  f , 
2,  i  of  J-  of  1^. 

4.  Add  4i,  1,  f  of  H,  3,  and  H. 

5.  Find  the  difference  between  f  of  6^  and  |  of  4,8^. 

Ans.  liff. 

6.  The  less  of  two  numbers  is  475 6|,  and  their  difference 
is  128|;  what  is  the  greater  number?  Ans.  4885^. 

7.  What  is  the  difference  between  the  continued  products 
of  3,  J,  I,  4|,  and  3^,  f ,  4,  f  ?  An^.  3^. 

4  2i 

8.  Reduce  the  fractions  -  and  -j  to  their  simplest  form. 

9.  What  number  multiplied  by  i  will  produce  1825 J  ? 

10.  A  farmer  had  ^  of  his  sheep  in  one  pasture,  J  in 
another,  and  the  remainder,  which  were  77,  in  a  third  pas- 
ture ;  how  many  sheep  had  he  ?  Ans.  140. 

11.  What  will  7i  cords  of  wood  cost  at  J  of  9^  dollars 
per  cord?  Ans.  $24^. 


PKOMISCUOUS    EXAMPLES.  113 

12.  At  I  of  a  dollar  per  bushel,  how  many  bushels  of  ap- 
ples can  be  bought  for  5-J  dollars  ? 

13.  Paid  $183  7|  for  7350 J  bushels  of  oats ;  how  much  was 
that  per  bushel  ?  .      Ans.  i  of  a  dollar. 

14.  If  235  J  acres  of  land  cost  $4725|,  what  will  628  acres 
cost  ?  Ans.  112601. 

15.  A  man,  owning  |  of  an  iron  foundry,  sold  ^  of  his 
share  for  $540| ;  what  was  the  value  of  the  foundry  ? 

Ans.  $4055|. 

16.  14f  less  ^-^^  is  f  of  I  of  what  number  ?  Ans.  27. 

17.  A  merchant  bought  4J  cords  of  wood  at  $3J  per  cord, 
and  paid  for  it  in  cloth  at  -|  of  a  dollar  per  yard ;  how  many 
yards  were  required  to  pay  for  the  wool  ? 

18.  How  many  yards  of  cloth,  f  of  a  yard  wide,  will  line 
20i  yards,  IJ  yards  wide  ?  Ans.  34J-. 

19.  If  the  dividend  be  |,  and  the  quotient  ^,  what  is 
the  divisor? 

20.  If  the  sum  of  two  fractions  be  |,  and  one  of  them  be 
^,  what  is  the  other  ?  A7is.  :^. 

21.  If  the  smaller  of  two  fractions  be  |f ,  and  their  differ- 
ence -^j  what  is  the  greater?  Ans.  -Jf. 

22.  If  3f  pounds  of  sugar  cost  33  cents,  what  must  be 
paid  for  65 J-  pounds  ? 

23.  If  324  bushels  of  barley  can  be  had  for  259^  bushels 
of  corn,  how  much  barley  can  be  had  for  2000  bushels  of 
com  ?  Ans.  2500  bushels. 

24.  A  certain  sum  of  money  is  to  be  divided  among  5  per- 
sons ;  A  is  to  have  ^J,  B  ^,  0  3^,  D  ^,  and  E  the  remain- 
der, which  is  20  dollars ;  what  is  the  whole  sum  to  be  di- 
Wded?  Ans.  $50. 

25.  What  number,  diminished  by  the  difference  between 
f  and  f  of  itself,  leaves  a  remainder  of  34  ?  Ans.  40. 

26.  If  I  of  a  farm  be  valued  at  $1728,  what  is  the  value 
of  the  whole  ? 


114  FEACTIOITS. 

27.  Bouglit  320  sheep  at  $2^-  per  head  ;  afterward  bought 
435  at  |1|  per  head  ;  then  sold  |  of  the  whole  number  at 
$1|  per  head,  and  the  remainder  at  $2|- ;  did  I  gain  or  lose, 
and  how  much  ?  Ans.  Lost  $44 J. 

28.  If  5  be  added  to  both  terms  of  the  fraction  -J,  will  its 
value  be  increased  or  diminished  ?       Ans.  Increased  ySt- 

29.  If  5  be  added  to  both  terms  of  the  fraction  f,  will  its 
value  be  increased  or  diminished  ?     Ans.  Diminished  -^. 

30.  How  many  times  can  a  bottle  holding  J  of  |  of  a 
gallon,  be  filled  from  a  demijohn  containing  |  of  1|  gallons  ? 

Ans.  7i. 

31.  Bought  i  of  7i  cords  of  wood  for  i  of  $32;  what  did 
1  cord  cost  ? 

•32.  Purchased  728  pounds  of  candles  at  16f  cents  a  pound; 
had  they  been  purchased  for  3-|  cents  less  a  pound,  how  many 
pounds  could  have  been  purchased  for  the  same  money  ? 

Ans.  953i|. 

33.  What  number,  divided  by  1-|,  will  give  a  quotient  of 
9i?  Ans.  12ii. 

34.  The  product  of  two  numbers  is  6,  and  one  of  them  is 
1846 ;  what  is  the  other  ?  Ans.  -^. 

35.  A  stone  mason  worked  llf  days,  and  after  paying 
his  board  and  other  expenses  with  f  of  his  earnings,  he  had 
$20  left ;  how  much  did  he  receive  a  day  ? 

36.  If  I  of  4  tons  of  coal  cost  $5|,  what  will  |  of  2  tons 
cost?  A71S.  $5. 

37.  In  an  orchard  f  of  the  trees  are  apple  trees,  -^  peach 
trees,  and  the  remainder  are  pear  trees,  which  are  20  morp 
than  4  of  the  whole ;  how  many  trees  in  the  orchard  ? 

Ans.  800. 

38.  A  man  gave  6-|  pounds  of  butter,  at  12  cents  a  pound, 
for  f  of  a  gallon  of  oil ;  what  was  the  oil  worth  a  gallon  ? 

Ans.  100  cents. 

39.  A:  gentleman,  having  2711-  acres  of  land,  sold  ^  of  it, 
and  gave  |  of  it  to  his  son  ;  what  was  the  value  of  the  re* 
mainder,  at  $57^  per  acre  ?  A?is.  $4577A- 


PEOMISCUOUS    EXAMPLES.  115 

40.  A  horse  and  wagon  cost  1270  ;  the  horse  cost  IJ  times 
as  much  as  the  wagon  ;  what  was  the  cost  of  the  wagon  ? 

41.  What  number  taken  from  2 J-  times  12|  will  leave 
20f?  J-ns.  11  J. 

42.  A  merchant  bought  a  cargo  of  flour  for  $2173|^,  and 
sold  it  for  If  of  the  cost,  thereby  losing  |  of  a  dollar  per 
barrel ;  how  many  barrels  did  he  purchase  ?      Ans.  126. 

43.  A  and  B  can  do  a  piece  of  work  in  14  days  ;  A  can  do 
J  as  much  as  B  ;  in  how  many  days  can  each  do  it  ? 

Ans,  A,  32f  days  ;  B,  24^  days. 

44.  How  many  yards  of  cloth  f  of  a  yard  wide,  are  equal 
to  12  yards  |  of  a  yard  wide  ?  Ans.  llj. 

45.  A,  B,  and  0  can  do  a  piece  of  work  in  5  days  ;  B  and 
C  can  do  it  in  8  days  ;  in  what  time  can  A  do  it  ? 

46.  A  man  put  his  money  into  4  packages  ;  in  the  first  ho 
put  f ,  in  the  second  ^,  in  the  third  -J-,  and  in  the  fourth  the 
remainder,  which  was  $24  more  than  -^  of  the  whole ;  how 
much  money  had  he  ?  Ans.  1720. 

47.  If  $7}  will  buy  3  J  cords  of  wood,  how  many  cords  can 
be  bought  for  110^  ?  Ans.  4f|. 

48.  How  many  times  is  ^  of  f  of  27  contained  in  -J  of  J  of 
42|? 

49.  A  boy  lost  J-  of  his  kite  string,  and  then  added  30 
feet,  when  it  was  just  ^  of  its  original  length  ;  what  was 
the  length  at  first  ?  Ans.  100  feet. 

50.  Bought  -f  of  a  box  of  candles,  and  having  used  -J  of 
them,  sold  the  remainder  for  -J^  of  a  dollar ;  how  much 
would  a  box  cost  at  the  same  rate  ?  Ans.  $5-^. 

51.  A  post  stands  ^  in  the  mud,  J  in  the  water,  and  21 
feet  above  the  water  ;  what  is  its  length  ? 

52.  A  father  left  his  eldest  son  f  of  his  estate,  his  youngest 
son  ^  of  the  remainder,  and  his  daughter  the  remainder, 
who  received  |1723|  less  than  the  youngest  son  ;  what  was 
the  value  of  the  estate?  Ans.  $21114J|. 


116  DECIMALS. 


DECIMAL   FEACTIONS. 

143.  Decimal  Fractions  are  fractions  which  have  for 
their  denominator  10,  100,  1000,  or  1  with  any  number  of 
ciphers  annexed. 

1.  The  word  decimal  is  derived  from  the  Latin  decern^  which  signifies  ten, 

2.  Decimal  fpactions  are  commonly  called  decimals. 

3.  Since  -^  =.  ^y^,  y^  =  yuwu*  ®*'^-»  ^^^  denominators  of  decimal  fractions  in- 
crease and  decrease  in  a  tenfold  ratio,  the  same  as  simple  numbers. 


Decimal  Notatiok  akd  Numebation. 

144,  Common  Fractions  are  the  common  divisions  of 
a  unit  into  any  number  of  equal  parts,  as  into  halves y  fifths, 
twenty-fourths,  etc. 

Decimal  Fractions  are  the  decimal  divisions  of  a  unit, 
thus  :  A  unit  is  divided  into  ten  equal  parts,  called  tenths  ; 
each  of  these  tenths  is  divided  into  ten  other  equal  parts, 
called  hundredths  ;  each  of  these  hundredths  into  ten  other 
equal  parts,  called  thousandths ;  and  so  on.  Since  the  de- 
nominators of  decimal  fractions  increase  and  decrease  by  the 
scale  of  10,  the  same  as  simple  numbers,  in  writing  decimals 
the  denominators  may  be  omitted. 

In  simple  numbers,  the  unit,  1,  is  the  starting  point  of 
notation  and  numeration  ;  and  so  also  is  it  in  decimals.  We 
extend  the  scale  of  notation  to  the  left  of  units'  place  in 
writing  integers,  and  to  the  right  of  units'  place  in  writing 
decimals.  Thus,  the  first  place  at  the  left  of  units  is  tens, 
and  the  first  place  at  the  right  of  units  is  tenths  ;  the  second 
place  at  the  left  is  hundreds,  and  the  second  place  at  the 
right  is  hundredths  ;  the  third  place  at  the  left  is  thousands, 
ftnd  the  third  place  at  the  right  is  thousandths ;  and  so  on. 

What  are  decimal  fractions?  How  do  they  diflFer  from  common 
fractions  ?    How  are  they  written  ? 


NOTATION    AND    NUMERATION.  117 

The  Decimal  Point  is  a  period  ( . ),  wliicli  must  always  be 
placed  before  or  at  the  left  hand  of  the  decimal.    Thus, 
-^    is  expressed  .6 

The  decimal  point  Is  also  called  the  Separatrix.    This  is  a  correct  name  for  it 
only  when  it  stands  between  the  integral  and  decimal  parts  of  the  same  numbei*. 

.5      is  5  tenths,  which  =  -jig-  of  5  units  ; 

.05    is  5  hundredths,         ''     z=z^^ofb  tenths  ; 
.005  is  5  thousandths,       '^     z=^ofh  hundredths. 
And  universally,  the  value  of  a  figure  in  any  decimal  place 
is  -^  the  value  of  the  same  figure  in  the  next  left  hand  place. 
The  relation  of  decimals  and  integers  to  each  other  is 
clearly  shown  by  the  following 

Numeration  Table. 


t 
s 

1 

. 

CD 

1 

i 

i 

.2 
1 

<K 

O 

1 

4 

1 

rn 

S 

1 

■d 

usandths 
-thousan 
dred-tho 

1 

1 
1 

'I 

i 

1 

^ 

1 

1 

w 

1 

1 

S   e^    w 

1 

9 

8 

7 

6 

5 

4 

3 

3 

1 

.2 

3 

4    5     6 

7 

8 

9 

Integers. 

Decimals. 

By  examining  this  table  we  see  that 

Tenths  are  expressed  by  one    figure. 

Hundredths  ''  ''         '^  two    figures. 

Thousandths         "  ''         "  three     '' 

Ten  thousandths  ''  "         ''  four      " 

And  any  order  of  decimals  by  one  figure  less  than  the  cor- 
responding order  of  integers. 

145.    Since  the  denominator  of  tenths  is  10,  of  hun- 

What  is  the  decimal  point  ?    What  is  it  sometimes  called  ?    Wha* 
is  the  value  of  a  figure  in  any  decimal  place  ? 


118  DECIMALS. 

dredths  100,  of  thousands  1000,  and  so  on,  a  decimal  may  be 
expressed  by  writing  the  numerator  only  ;  but  in  this  case 
the  numerator  or  decimal  must  always  contain  as  many 
decimal  places  as  are  equal  to  the  number  of  ciphers  in  the 
denominator ;  and  the  denominator  of  a  decimal  will  always 
be  the  unit,  1,  with  as  many  ciphers  annexed  as  are  equal  to 
the  number  of  figures  in  the  decimal  or  numerator. 
The  decimal  point  must  never  be  omitted. 

Examples  for  Peactice. 

1.  Express  in  figures  thirty-eight  hundredths. 

2.  Write  seven  tenths. 

3.  Write  three  hundred  twenty-five  thousandths. 

4.  Write  four  hundredths.  A?is,  .04. 

5.  Write  sixteen  thousandths. 

6.  Write  seventy-four  hundred-thousandths.  Ans.  .00074. 

7.  Write  seven  hundred  forty-five  millionths. 

8.  Write  four  thousand  two  hundred  thirty-two  ten-thou- 
sandths. 

9.  Write  five  hundred  thousand  millionths. 
10.  Eead  the  following  decimals : 

.05  .681  .9034  .19248 

.24  .024  .0005  .001385 

.672  .8471  .100248  .1000087 

To  read  a  decimal,  first  numerate  from  left  to  right,  and  the  name  of  the  right 
hand  figure  is  the  name  of  the  denominator.  Then  numerate  from  right  to  left,  aa 
in  whole  numbers,  to  read  the  numerator. 

146,  A  Mixed  Number  is  a  number  consisting  of  in- 
tegers and  decimals ;  thus,  71.406  consists  of  the  integral 
part,  71,  and  the  decimal  part,  .406 ;  it  is  read  the  same  aa 
■^ItWV*  '^1  and  406  thousandths. 

Examples  for  Practice. 

1.  Write  eighteen,  and  twenty-seven  thousandths. 

2.  Write  four  hundred,  and  nineteen  ten-millionth s. 

How  many  decimal  places  must  there  be  to  express  any  decimal  ? 


NOTATIOJ?^    AND    NUMERATION.  119 

3.  Write  fifty-four,  and  fifty-four  millionths. 

4.  Eighty-one,  and  1  ten-thousandth. 

5.  One  hundred,  and  67  ten-thousandths. 

6.  Bead  the  following  numbers  : 

18.027  100.0067  400.0000019 

81.0001  64.000054  3.03 

75.075  9.2806  40.40404 

147.  Trom  the  foregoing  explanations  and  illustrations 
we  derive  the  following  important 

Peinciples  of  Decimal  Notation  and  !N"umeration. 

1.  The  value  of  any  decimal  figure  depends  upon  its  place 
from  the  decimal  point :  thus  .3  is  ten  times  .03. 

2.  Prefixing  a  cipher  to  a  decimal  decreases  its  value  the 
same  as  dividing  it  by  ten  ;  thus,  .03  is  ^  the  value  of  .3. 

3.  Annexing  a  cipher  to  a  decimal  does  not  alter  its  value, 
since  it  does  not  change  the  place  of  the  significant  figures 
of  the  decimal ;  thus,  -^^  or  .6,  is  the  same  as  3%,  or  .60. 

4.  Decimals  increase  from  right  to  left,  and  decrease  from 
left  to  right,  in  a  tenfold  ratio  ;  and  therefore  they  may  be 
added,  subtracted,  multiplied,  and  divided  the  same  as  whole 
numbers. 

5.  The  denominator  of  a  decimal,  though  never  expressed, 
is  always  the  unit,  1,  with  as  many  ciphers  annexed  as  there 
are  figures  in  the  decimal. 

6.  To  read  decimals  requires  two  numerations ;  first,  from 
units,  to  find  the  name  of  the  denominator,  and  second,  tO' 
wards  units,  to  find  the  value  of  the  numerator. 

148.  Having  analyzed  all  the  principles  upon  which  the 
writing  and  reading  of  decimals  depend,  we  will  now  pre- 
sent these  principles  in  the  form  of  rules. 

EuLE  FOR  Decimal  Notation. 
I.   Write  the  decimal  the  same  as  a  ivhole  number,  placing 

What  is  the  first  principle  of  decimal  notation  ?  Second  ?  Third  1 
Fourth?    Fifth?    Sixth?    Rule  for  notation,  first  step? 


120  DECIMALS. 

ciphers  where  necessary  to  give  each  significant  figure  its  true 
local  value, 
IL  Place  the  decimal  point  before  the  first  figure. 

EuLE  FOR  Decimal  Numeration. 

I.  Numerate  frmn  the  decimal  point,  to  determine  the  de- 
nominator, 

II.  Numerate  towards  the  decimal  point,  to  determine  the 
numerator. 

III.  Bead  the  decimal  as  a  whole  number,  giving  it  the 
name  or  denomination  of  the  right  hand  figure. 

Examples  for  Practice. 

1.  Write  425  millionths. 

2.  Write  six  thousand  ten-thousandths. 

3.  Write  one  thousand  eight  hundred  fif  tj-nine  hundred- 
thousandths. 

4.  Write  260  thousand  8  billionths. 

5.  Eead  the  following  decimals  : 

.6321  .748243  .2902999 

.5400027  .60000000  .00000006 

6.  Write  five  hundred  two,  and  one  thousand  six  mil- 
lionths. 

7.  Write  thirty-one,  and  two  ten-millionths. 

8.  Write  eleven   thousand,  and  eleven  hundred-thou- 
Bandths 

9.  Write  nine  million,  and  nine  billionths. 

10.  Write  one  hundred  two  tenths.  Ans,  10.2. 

11.  Write  one  hundred  twenty-four  thousand  three  hun- 
dred fifteen  thousandths. 

12.  Write  seven  hundred  thousandths. 

13.  Write  seven  hundred-thousandths. 

14.  Read  the  following  numbers  : 

12.36  9.052  62.9999 

142.847  32.004  1858.4583 

1.02  4.0005  27.00045 

Second  ?    Rule  for  numeration,  first  step  ?    Second  ?    Third  ? 


EEDUCTION.  121 

EEDUCTION. 

Case  I. 

149.  To  reduce  decimals  to  a  common  denomina- 
tor. 

1.  Eeduce  .5,  .375,  3.25401,  and  46.13  to  their  least  com- 
mon decimal  denominator. 

OPEEATiON.  Analysis.    The  given  decimals  must  contain 

,50000  as  many  places  each,  as  are  equal  to  the  greatest 

37500  number  of  decimal  figures  in  any  of  the  given 

onKAoi  decimals.    We  find  that  the  third  number  con- 

tains five  decimal  places,  and  hence  100000  must 
%0,lo\j0\)  jjg  ^  common  denominator.    As  annexing  ciphers 

to  decimals  does  not  alter  their  value,  (14:4,  3) 
we  give  to  each  number  five  decimal  places  by  annexing  ciphers,  and 
thus  reduce  the  given  decimals  to  a  common  denominator. 

KuLE.  Give  to  each  number  the  same  numler  of  decimal 
places,  ly  annexing  ciphers, 

1.  If  the  numbers  be  reduced  to  the  denominator  of  that  one  of  the  given  num- 
bers having  the  greatest  number  of  decimal  places,  they  will  have  their  least  com- 
mon decimal  denominator. 

2.  A  whole  number  may  readily  be  reduced  to  decimals  by  placing  the  decimal 
point  after  units,  and  annexing  ciphers ;  one  cipher  reducing  it  to  tenths,  two 
ciphers  to  hundredths^  three  ciphers  to  thousandths,  and  so  on. 

Examples  fob  Practice. 

2.  Eeduce  .17,  24.6,  .0003,  84,  and  721.8000271  to  their 
least  common  denominator. 

3.  Eeduce  7  tenths,  24  thousandths,  187  millionths,  5 
hundred  millionths,  and  10845  hundredths  to  their  least 
common  denominator. 

4.  Eeduce  to  their  least  common  denominator  the  following 
decimals  :  1000.001,  841.78,  2.6004,  90.000009,  and  6000. 

What  is  meant  by  the  reduction  of  decimals?    Case  I  is  what? 
Give  explanation.    Eule. 
E.  P.  6 


Ans. 

h 

Am. 

A- 

Ans. 

m- 

Ans. 

ih 

122  decimals. 

Case  IL 

150.  To  reduce  a  decimal  to  a  common  fraction. 

1.  Reduce  .75  to  ifcs  equivalent  common  fraction. 

Analysis.    Omit  the  decimal  point,  supply 
OPERATION.  the  proper  denominator  to  the  decimal,  and 

.75=:3^=|.  then  reduce  the  common  fraction  thus  formed 

to  its  lowest  terms. 

EuLE.   Oinit  the  decimal  point,  and  supply  the  proper  de- 
nominator. 

Examples  foe  Peactice. 

2.  Eeduce  .125  to  a  common  fraction. 

3.  Reduce  .16  to  a  common  fraction. 

4.  Reduce  .655  to  a  common  fraction. 

5.  Reduce  .9375  to  a  common  fraction. 

6.  Reduce  .0008  to  a  common  fraction.  Ans,  -rhu- 

Case  III. 

151.  To  reduce  a  common  fraction  to  a  decimal. 

1.  Reduce  J  to  its  equivalent  decimal. 

FIRST  OPERATION.  ANALYSIS.     First  annex  the 

^:^^^z=zr^^=z.'l[5,  Ans,  same    number    of    ciphers   to 

both    terms  of   the    fraction; 
SECOND  OPERATION.  this  does  not  alter  its  value. 

4)3.00  Then    divide    both    resulting 

~  terms  by  4,  the  significant  fig- 

ure of  the  denominator,  to  ob- 
tain the  decimal  denominator, 
100.    Then  the  fraction  is  changed  to  the  decimal  form  by  omitting 
the  denominator.      If  the  intermediate  steps  be  omitted,  the  true 
result  may  be  obtained  as  in  the  second  operation. 

2.  Reduce  ^  to  its  equivalent  decimal. 

Case  II  is  what  ?    Give  explanation.    Rule.    Case  III  is  what  ?    Ex- 
plain  first  operation.    Second. 


KEDUCTIOK, 


123 


THIRD  OPERATION. 

16  )  1.0000 

.0625,  Ans. 


Analysis.  Dividing  as  in  the  former 
example,  we  obtain  a  quotient  of  3  fig- 
ures, 625.  But  since  we  annexed  4 
ciphers,  there  must  be  4  places  in  the 


required  decimal;    hence  we   prefix  1  cipher.     This  is  made  still 
plainer  by  the  following  operation  ;  thus, 

i^=iVWft=TtUo=.0625. 

From  these  illustrations  we  deriye  the  following 

EuLE.  I.  Annex  ciphers  to  the  numerator,  and  divide  by 

the  denominator. 
II.  Point  off  as  many  decimal  places  in  the  result  as  are 

equal  to  the  number  of  ciphers  annexed. 

A  common  fraction  can  be  reduced  to  an  exact  decimal  when  its  lowest  denomi- 
nator contains  only  the  prime  factors  2  and  5,  and  not  otherwise. 

Examples  foe  Peactice. 


3.  Eeduce  |  to  a  decimal. 

Ans. 

.625. 

4.  Reduce  f  to  a  decimal. 

5.  Reduce  ^  to  a  decimal. 

Ans. 

.9375. 

6.  Reduce  -J  to  a  decimal. 

7.  Reduce  -^  to  a  decimal. 

Ans. 

.08. 

8.  Reduce  -^  to  a  decimal. 

Ans. 

.046875. 

9.  Reduce  -|  to  a  decimal. 

10.  Reduce  -^^  to  a  decimal. 

11.  Reduce  -^  to  a  decimal 

Ans. 

.00375. 

12.  Reduce  y|^  to  a  decimal. 

Ans. 

.008. 

13.  Reduce  -J-  to  a  decimal. 

Ans. .; 

33333  +  . 

The  sign,  +,  in  the  answer  indicates  that  there  is  still  a  remainder. 

14.  Reduce  ^  to  a  decimal.  Ans.  .513513  +  . 

The  answers  to  the  last  two  examples  are  called  repeating  decimals  ;  and  the 
figure  3  in  the  13th  example,  and  the  figures  513  in  the  14th,  are  called  repetends, 
because  they  are  repeated,  or  occur  in  regular  order. 


Third  operation.    Rule,  first  step ?    Second?    When  can  a  common 
fraction  be  reduced  to  an  exact  decimal  ? 


124  DECIMALS. 


ADDITION. 

152.  1.  What  is  the  sum  of  3.703,  621.57,  .672,  and 
20.0074? 

OPERATION.  Analysis.    Write  the  numbers  so  that  figures 

3.703  of  like  orders  of  units  shall  stand  in  the  same 

g21  57  columns ;  that  is,  units  under  units,  tenths  un- 

^-,jj  der  tenths,  hundredths  under  hundredths,  etc. 

This  brings  the  decimal  points  directly  under 

20.0074  each  other.     Commencing  at  the  right  hand,  add 

645.9524  each  column  separately,  and  carry  as  in  whole 

numbers,  and  in  the  result  place  a  decimal  point 

between  units  and  tenths,  or  directly  under  the  decimal  point  in  the 

numbers  added. 

EuLE.  I.  Write  the  numbers  so  that  the  decimal  points 
shall  stand  directly  under  each  other. 

II.  Add  as  in  whole  numbers,  and  place  the  decimal  point, 
in  the  result,  directly  under  the  points  in  the  numbers  added. 

Examples  for  Practice. 

2.  Add  .199  3.  Add  4.015 

2.7569  6.75 

.25  27.38203 

.654  375.01 

Sum,  3.8599  ^-^ 

Amount,  415.65703 

4.  Add  1152.01,  14.11018, 152348.21,  9.000083. 

Ans.  153523.330263. 

5.  Add  37.03,  0.521,  .9,  1000,  4000.0004. 

Ans.  5038.4514. 

6.  What  is  the  sum  of  twenty-six,  and  twenty-six  hun- 
dredths ;  seven  tenths ;  six,  and  eighty-three  thousandths ; 
four,  and  four  thousandths?  Ans.  37.047. 

Explain  the  operation  of  addition  of  decimals.  Qive  rule,  first  step. 
Second. 


ADDITIOIT.  125 

7.  What  is  the  sum  of  thirty-six,  and  fifteen  thousandths  ; 
three  hundred,  and  six  hundred  five  ten-thousandths ;  five, 
and  three  million  fchs;  sixty,  and  eighty-seven  ten-millionths  ? 

Ans.  401.0755117. 

8.  What  is  the  sum  of  fifty-four,  and  thirty-four  hun- 
dredths; one,  and  nine  ten-thousandths;  three,  and  two 
hundred  seven  millionths ;  twenty-three  thousandths ;  eight, 
and  nine  tenths ;  four,  and  one  hundred  thirty-five  thou- 
sandths? Ans.n.dddlOH. 

9.  How  many  yards  in  three  pieces  of  cloth,  the  first  piece 
containing  18.375  yards,  the  second  piece  41.625  yards,  and 
the  third  piece  35.5  yards  ? 

10.  A's  farm  contains  61.843  acres,  B's  contains  143.75 
acres,  O's  218.4375  acres,  and  D's  21.9  acres;  how  many 
acres  in  the  four  farms  ? 

11.  My  farm  consists  of  7  fields,  containing  12|  acres, 
18f  acres,  9  acres,  24^  acres,  4^  acres,  8^  acres,  and  15J4 
acres  respectively ;  how  many  acres  in  my  farm  ? 

Reduce  the  common  fractions  to  decimals  before  adding. 

A71S,  93.6375. 

12.  A  grocer  has  2^  barrels  of  A  sugar,  5f  barrels  of  B 
sugar,  3f  barrels  of  C  sugar,  3.0642  barrels  of  crushed 
sugar,  and  8.925  barrels  of  pulverized  sugar ;  how  many 
barrels  of  sugar  has  he  ?  Ans»  23.8642. 

13.  A  tailor  made  3  suits  of  clothes  ;  for  the  first  suit  he 
used  2-J  yards  of  broadcloth,  3^  yards  of  cassimere,  and  | 
yards  of  satin  ;  for  the  second  suit  2.25  yards  of  broadcloth, 
2.875  yards  of  cassimere,  and  1  yard  of  satin  ;  and  for  the 
third  suit  5-f^  yards  of  broadcloth,  and  1-J-  yards  of  satin. 
How  many  yards  of  each  kind  of  goods  did  he  use  ?  How 
many-yards  of  all?  Ans.  to  last,  18.375. 


126 


DECIMALS. 


SUBTEACTION, 


153.  1.  From  91.73  take  2.18. 

OPERATION. 
91.73 

2.18 


Ans.  89.55 
2.  From  2.9185  take  1.42. 

OPERATION. 

2.9185 
1.42 


Ans.  1.4985 
3.  From  124.65  take  95.58746. 

OPERATION. 

124.65 
95.58746 
Ans.  29.06254 


Analysis.  In  each  of  these 
three  examples,  we  write  the 
subtrahend  under  the  mm- 
uend,  placing  units  under 
units,  tenths  under  tenths, 
etc.  Commencing  at  the 
right  hand,  we  subtract  as 
in  whole  numbers,  and  in 
the  remainders  we  place  the 
decimal  points  directly  under 
those  in  the  numbers  above. 
In  the  second  example,  the 
number  of  decimal  places  in 
the  minuend  is  greater  than 
the  number  in  the  subtra- 
hend, and  in  the  third  exam- 
ple the  number  is  less.  In 
both  cases,  we  reduce  both 
minuend  and  subtrahend  to 
the  same  number  of  decimal 
places,  by  annexing  ciphers; 
or  Ave  suppose  the  ciphers  to 


be  annexed,  before  performing  the  subtraction. 

Rule.  I.  Write  the  7ium'bers  so  that  the  decimal  j)oi7its 
shall  stand  directly  under  each  other. 

II.  Subtract  as  in  whole  numbers,  and  place  the  decimal 
point  in  the  result  directly  under  the  points  in  the  given 
numbers, 

4.  Find  the  difference  between  714  and  .916.  Ans.  713.084. 

5.  How  much  greater  is  2  than  .298  ?  Am.  1.702. 

6.  From  21.004  take  75  hundredths. 

7.  From  10.0302  take  2  ten-thousandths.      Ans,  10.03. 

8.  From  900  take  .009.  Ans.  899.991. 

9.  From  two  thousand  take  two  thousandths. 

10.  From  one  take  one  millionth.  Ans.  .990999. 


Expl.iin  subtraction  of  fractions.     Give  the  rule,  first  step.     Second. 


I 


MULTIPLICATION.  127 

11.  From  four  hundred  twenty-seven  thousandths  take 
four  hundred  twenty-seven  milHonths.         Ans.  .426573. 

12.  A  man  owned  thirty-four  hundredths  of  a  township 
of  land,  and  sold  thirty-four  thousandths  of  the  township  ; 
how  much  did  he  still  own  ?  Ans.  .306. 


MULTIPLICATION. 

154.   1.  What  is  the  product  of  .35  multiplied  by  ,5  ? 

OPERATION.  Akalysis,     We  perform  the  multiplication  the 

.35  same  as  in  whole  numbers,  and  the  only  difficulty 

5  we  meet  with  is  in  pointing  off  the  decimal  places 

in  the  product     To  determine  how  many  places  to 

.175,  A?IS.  point  off,  we  may  reduce  the  decimals  to  common 
fractions ;  thus,  .35  =  -^^^  and  .5  —  /^.  Perform- 
ing the  multiplication,  we  have  ^^\  x  tu  =  tuwg'  ^^^  t^^is  product, 
expressed  decimally,  is  .175.  Here  we  see  that  the  product  contains 
as  many  decimal  places  as  are  contained  in  both  multiplicand  and 
multiplier. 

EuLE.  Multiply  as  in  whole  numlers,  and  from  the  right 
hand  of  the  product  point  off  as  many  figures  for  decimals  as 
there  are  decimal  places  in  both  factors. 

1.  If  there  be  not  as  many  figures  in  the  product  as  there  are  decimals  in  both  fiwj- 
tors,  supply  the  deficiency  by  prefixing  ciphers. 

2.  To  multiply  a  decimal  by  10, 100, 1000,  etc.,  remove  the  point  as  many  places  to 
♦he  right  as  there  are  ciphers  on  the  right  of  the  multiplier. 

Examples  for  Practice. 

"Z.  Multiply  1.245  by  .27.  Ans.  .33615. 

3.  Multiply  79.347  by  23.15.  Ans.  1836.88305. 

4.  Multiply  350  by  .7853. 

5.  Multiply  one  tenth  by  one  tenth.  Ans.  .01. 

6.  Multiply  25  by  twenty-five  hundredths.      Ans.  6.25. 

Explain  multiplication  of  decimals.  Give  rule.  If  the  product 
have  less  decimal  places  than  both  factors,  how  proceed  ?  How  mul- 
tiply by  10,  100,  1000,  etc? 


t28  DECIMALS. 

7.  Multiply  .132  by  .241.  Ans,  .031812. 

8.  Multiply  24.35  by  10. 

9.  Multiply  .006  by  1000.  Ans,  6. 

10.  Multiply  .23  by  .009.  Ans.    .00207. 

11.  Multiply  sixty-four  tbousandths  by  thirteen  mil- 
liontlis.  Ans,  .000000832. 

12.  Multiply  eigbty-seTen  ten-thousandths  by  three  hun- 
dred fifty-two  hundred-thousandths. 

13.  Multiply  one  million  by  one  millionth.         Ans,  1, 

14.  Multiply  sixteen  thousand  by  sixteen  ten-thousandths. 

Ans.  25.6. 

15.  If  a  cord  of  wood  be  worth  2.37  bushels  of  wheat, 
how  many  bushels  of  wheat  must  be  given  for  9.58  cords  of 
wood?  Ans,  22.7046  bushels. 


DIVISION. 

155.   1.  What  is  the  quotient  of  .175  divided  by  .5  ? 

OPEKATION.         Analysis.    Performing  the  division  the  same  as 

.5  )  .175       in  whole  numbers,  the  only  diflBculty  we  meet  with 

J  rr~       is  in  pointing  oflf  the  decimal  places  in  the  quotient. 

*    *  To  determine  how  many  places  to  point  off,  we  may 

reduce  the  decimals  to  common   fractions ;  thus,  .175  =  xVin7>  ^^^ 

J5=^.    Performing  the  division,  we  have 

35 
175    5    m       10   35 


X  -^  =  T7^  ; 


1000  *  10   1000   $        100 

»nd  this  quotient,  expressed  decimally,  is  .35.  Here  we  see  that  the 
dividend  contains  as  many  decimal  places  as  are  contained  in  both 
divisor  and  quotient. 

Rule.  Divide  as  in  whole  numbers,  and  from  the  right 

hand  of  the  quotient  point  off  as  many  places  for  decimals 
(IS  the  decimal  places  in  the  dividend  exceed  those  in  the 
divisor. 

Explain  division  of  decimals.    Give  rule. 


DIVISION".  129 

1.  If  the  number  of  flgnres  in  the  quotient  be  less  than  the  excess  of  the  decimal 
places  in  the  dividend  over  those  in  the  divisor,  the  deficiency  must  be  supplied  by 
prefixing  ciphers. 

2.  If  there  be  a  remainder  after  dividing  the  dividend,  annex  ciphers,  and  continue 
the  division ;  the  ciphers  annexed  are  decimals  of  the  dividend. 

3.  The  dividend  must  always  contain  at  least  as  many  decimal  places  as  the  diyi- 
gor,  before  commencing  the  division. 

4.  In  most  business  transactions,  the  division  is  considered  sufliciently  exact 
when  the  quotient  is  carried  to  4  decimal  places,  unless  great  accuracy  is  required. 

5.  To  divide  by  10, 100, 1000,  etc.,  remove  the  decimal  point  as  many  places  to  the 
left  as  there  are  ciphers  on  the  right  hand  of  the  divisor. 


Examples  for  Practice. 

2.  Divide  .675  by  .15.  Ans.      4.5. 

3.  Divide  .288  by  3.6.  Ans,      .08. 

4.  Divide  81.6  by  2.5.  Ans,  32.64. 

5.  Divide  2.3421  by  21.1. 

6.  Divide  2.3421  by  .211. 

7.  Divide  8.297496  by  .153.  Ans.  54.232. 

8.  Divide  12  by  .7854. 

9.  Divide  3  by  3  ;  divide  3  by  .3  ;  3  by  .03  ;  30  by  .03. 

10.  Divide  15.34  by  2.7. 

11.  Divide  .1  by  .7.  Ans.  .142857  +  . 

12.  Divide  45.30  by  .015.  Ans.  3020. 

13.  Divide  .003753  by  625.5.  Ans.      .000006. 

14.  Divide  9  by  450.  Ans,  .0^. 

15.  Divide  2.39015  by  .007.  Ans.         341.45. 

16.  Divide  fifteen,  and  eight  hundred  seventy-five  thou- 
sandths, by  twenty-five  ten-thousandths.         Ans,    6350. 

17.  Divide  365  by  100. 

18.  Divide  785.4  by  1000.  Ans.   .7854. 

19.  Divide  one  thousand  by  one  thousandth. 

A71S.  1000000. 


When  are  ciphers  prefixed  to  the  quotient  ?  If  there  be  a  remainder, 
how  proceed  ?  If  the  dividend  have  less  decimal  places  than  the  divi- 
sor, how  proceed  ?    How  divide  by  10,  100,  1000,  etc.  ? 

6* 


130  decimals. 

Pkomiscuous  Examples. 

1.  Add  six  hundred,  and  twenty-five  thousandths ;  four 
tenths  ;  seven,  and  sixty-two  ten- thousandths ;  three,  and 
fifty-eight  millionth  s  ;  ninety-two,  and  seven  hundredths. 

Ans.  702.501258. 

2.  What  is  the  sum  of  81.003  -f-  5000.4  +  5.0008  -^ 
73.87563  +  1000  +  25  +  3.000548  +  .0315  ? 

3.  Prom  eighty-seven  take  eighty-seven  thousandths. 

4.  What  is  the  difference  between  nine  million  and  nine 
millionths  ?  Ans.  8999999.993991. 

5.  Multiply  .365  by  .15.  Ans.  .05475. 

6.  Multiply  three  thousandths  by  four  hundredths. 

7.  If  one  acre  produce  42.57  bushels  of  com,  how  many 
bushels  will  18.73  acres  produce  ?  Ans.  797.3361. 

8.  Divide  .125  by  8000.  Ans.  .000015625. 

9.  Divide  .7744  by  .1936. 

10.  Divide  27.1  by  100000.  Ans.  .000271. 

11.  If  6.35  acres  produce  70.6755  bushels  of  wheat,  what 
does  one  acre  produce  ?  Ans.  11.13  bushels. 

12.  Eeduce  .625  to  a  common  fraction.  Ans.  -|. 

13.  Express  26.875  by  an  integer  and  a  common  fraction. 

Ans.   26|. 

14.  Eeduce  yIt  ^^  ^  decimal  fraction.  Ans.  .016. 

15.  Eeduce  -—  to  a  decimal  fraction.  Ans.  .5. 

16.  How  many  times  will  .5  of  1.75  be  contained  in  .25  of 
17i?  Ans.  5. 

17.  What  will  be  the  cost  of  3|  bales  of  cloth,  each  bale 
containing  36.75  yards,  at  .85  dollars  per  yard  ? 

18.  Traveling  at  the  rate  of  4|  miles  an  hour,  how  many 
hours  will  a  man  require  to  travel  56.925  miles  ? 

Ans.  12|  hours. 


NOTATION     AND     NUMEEATION.  131 


DECIMAL    OUEEElSrOT. 

156.  Coin  is  money  stamped,  and  has  a  given  value  es- 
tablished by  law. 

157.  Currency  is  coin,  bank  bills,  treasury  notes,  etc., 
in  circulation  as  a  medium  of  trade. 

158.  A  Decimal  Currency  is  a  currency  whose  denom- 
inations increase  and  decrease  in  a  tenfold  ratio. 

The  currency  of  the  United  States  is  decimal  currency,  and  is  sometimes  called 
Federal  Money  ;  it  was  adopted  by  Congress  in  1786. 

NOTATION    AND    NUMERATION. 

BY  THE  "  COINAGE  ACT  OP  1873." 

The  €oin  of  the  United  States  consists  of  gold,  silver, 
nickel,  and  bronze. 

Tlie  Gold  Coifis  are  the  double-eagle,  eagle,  half-eagle, 
quarter-eagle,  three-dollar  and  one-dollar  pieces. 

Tlie  Silver  Coins  are  the  dollar,  half-dollar,  quarter-dol- 
lar, the  twenty-cent,  and  ten -cent  pieces. 

The  Mckel  Coins  are  the  five-cent  and  three-cent  pieces. 

The  Bronze  Coins  are  the  one-cent  pieces. 

Table. 

10  mills  (m.)  make  1  cent c 

10  cents  '^  1  dime d. 

10  dimes  "  1  dollar $. 

10  dollars  ''  1  eagle E. 

1.  The  mill  is  a  denomination  used  only  in  computations ;  it  is  not  a  coin. 

2.  The  trade-dollar  is  designed  solely  for  commerce,  and  not  for  currency.  Its 
weight  Is  420  grains.    The  weight  of  the  currency  dollar  of  1878  is  412^  grains. 

S.  The  character  $  is  supposed  to  be  a  contraction  of  U.  S.  (United  States),  the 
U  being  placed  upon  the  S. 

Wliat  is  coin?  Currency?  Decimal  currency?  Federal  money? 
What  are  the  gold  coins  of  U.  S.  ?  Silver  ?  Copper  ?  What  are  the 
denominations  of  U.  S.  currency  ?  What  is  the  sign  of  dollars  ?  From 
what  de jived  ? 


132  DECIMAL    CURRENCY. 

159.  The  gold  dollar  is  the  unit  of  United  States  money ; 
dimes,  cents,  and  mills  are  fractions  of  a  dollar,  and  are 
separated  from  the  dollar  by  the  decimal  point ;  thus,  two 
dollars  one  dime  two  cents  five  mills,  are  written  $2,125. 

By  examining  the  tabUy  we  see  that  the  dime  is  a  tenth 
part  of  the  unit,  or  dollar;  the  cent  a  tenth  part  of  the  dime 
or  a  hundredth  part  of  the  dollar  ;  and  the  mill  a  tenth  part 
of  the  cent,  a  hundredth  part  of  the  dime,  or  a  thousandth 
oart  of  the  dollar.  Hence  the  denominations  of  decimal 
currency  increase  and  decrease  the  same  as  decimal  fractions, 
and  are  expressed  according  to  the  same  decimal  system  of 
notation ;  and  they  may  be  added,  subtracted,  multiplied, 
and  divided  in  the  same  manner  as  decimals. 

Dimes  are  not  read  as  dimes,  but  the  two  places  of  dimes 
and  cents  are  appropriated  to  cents  ;  thus,  1  dollar  3  dimes 
2  cents,  or  $1.32,  are  read  one  dollar  thirty-two  cents  ;  hence. 

When  the  number  of  cents  is  less  than  10,  we  write  a 
cipher  before  it  in  the  place  of  dimes. 

The  half  cent  is  frequently  written  as  5  mills ;  thus,  24^  cents,  written  $.246. 

160.  Business  men  frequently  write  cents  as  common 
fractions  of  a  dollar  ;  thus,  three  dollars  thirteen  cents  are 
written  ^3^^,  and  read,  three  and  thirteen  hundredths 
dollars.  In  business  transactions,  when  the  final  result  of  a 
computation  contains  5  mills  or  more,  they  are  called  one 
cent,  and  when  less  than  5,  they  are  rejected. 

Examples  for  Practice. 

1.  Write  four  dollars  five  cents.  Ans.  $4.05. 

2.  Write  two  dollars  nine  cents. 

3.  Write  ten  dollars  ten  cents. 

.    4.  Write  eight  dollars  seven  mills.  Ans.  $8,007. 

What  is  the  unit  of  U.  S.  currency  ?  What  is  the  general  law  of 
increase  and  decrease  ?  In  practice,  how  many  decimal  places  are  given 
to  cents  ?  In  business  transactions,  how  are  cents  frequently  written  1 
WTiat  is  done  if  the  mills  exceed  5  ?    If  less  than  5  ? 


REDUCTION.  133 

5.  Write  sixty-four  cents.  Ans.  $0.64 

6.  Write  three  cents  two  mills. 

7.  Write  one  hundred  dollars  one  cent  one  milL 

8.  Read  $7.93;  $8.02;  $6,542. 

9.  Read  $5,272;  $100,025;  $17,005. 

10.  Read  $16,205;  $215,081;  $1000.011;  $4,002. 

REDUCTION. 

161.  By  examining  the  table  of  Decimal  Currency,  we 
see  that  10  mills  make  one  cent,  and  100  cents,  or  1000 
mills,  make  one  dollar  ;  hence, 

To  change  dollars  to  cents,  multiply  hy  100;  that  is,  annex 
two  cipliers. 

To  change  dollars  to  mills,  annex  three  ciphers. 
To  change  cents  to  mills,  annex  one  cipher. 

Examples  fob  Peactice. 

1.  Change  $792  to  cents.  Ans.  79200  cents. 

2.  Change  $36  to  cents. 

3.  Reduce  $5248  to  cents. 

4.  In  6.25  dollars  how  many  cents?        Ans.  625  cents. 

To  change  dollars  and  cents  to  cents,  or  dollars,  cents,  and  mills  to  mills,  remove 
the  decimal  point  and  the  sign,  $. 

5.  Change  $63,045  to  mills.  Ans,  63045  mills. 

6.  Change  16  cents  to  mills. 

7.  Reduce  $3,008  to  mills. 

8.  In  89  cents  how  many  mills  ? 

162.  Conversely, 

To  change  cents  to  dollars,  divide  ty  100;  that  is,  point 
off  two  figures  from  the  right. 

To  change  mills  to  dollars,  point  off  three  figures. 
To  change  mills  to  cents,  point  off  one  figure. 

How  are  dollars  changed  to  cents  ?  to  mills  ?  How  are  cents  changed 
to  mills  ?   How  are  cents  changed  to  dollars  ?  Mills  to  dollars  ?  to  cents  t 


134  decimal  cukrency. 

Examples  foe  Practice. 

1.  Change  875  cents  to  dollars.  Ans,  18.75. 

2.  Change  1504  cents  to  dollars. 

3.  In  13875  cents  how  many  dollars  ? 

4.  In  16525  mills  how  many  dollars  ? 

5.  Reduce  524  mills  to  cents. 

6   Eeduce  6524  mills  to  dollars. 

ADDITION. 

163.  1.  A  man  bought  a  cow  for  21  dollars  50  cents,  a 
horse  for  125  dollars  37J-  cents,  a  harness  for  46  dollars  75 
cents,  and  a  carriage  for  210  dollars  ;  how  much  did  he  pay 
for  all  ? 

OPERATION. 

$     21.50  Analysis:    Writing  dollars  under  dol- 

125.375  lars,  cents  under  cents,  etc.,  so  tliat  the 

^Q  WK  decimal  points  shall   stand  under  each 

rt-j  r.  ri/%  other,  add  and  point  off  as  in  addition  of 


Ans.  1403.625 


decimals. 


Rule.    I.  Write  dollars  under  dollars,  cents  under  ce7its,  etc. 
11.  Add  as  in  simple  numhers,  and  place  the  point  in  the 
amount  as  in  addition  of  decimals. 

Examples  for  Practice. 

2.  What- is  the  sum  of  50  dollars  7  cents,  1000  dollars  75 
cents,  60  dollars  3  mills,  18  cents  4  mills,  1  dollar  1  cent, 
and  25  dollars  45  cents  8  mills  ?  Ans.  $1137.475. 

3.  Add  364  dollars  54  cents  1  mill,  486  dollars  6  cents,  93 
dollars  9  mills,  1742  dollars  80  cents,  3  dollars  27  cents  6 
mills.  A71S.  $2689.686. 

4.  Add  92  cents,  10  cents  4  mills,  35  cents  7  mills,  18  cents 
6  mills,  44  cents  4  mills,  12|-  cents,  and  99  cents.  Ans.  $3,126. 

Explain  the  process  of  addition  of  decimal  currency.  Rule,  first 
Btep.    Second. 


SUBTEACTIOlir.  135 

5.  A  farmer  receives  89  dollars  74  cents  for  wheat,  13 
dollars  3  cents  for  corn,  6  dollars  37^  cents  for  potatoes, 
and  19  dollars  62 J  cents  for  oats;  what  does  he  receive  for 
the  whole?  Ans.  ^US.H'T. 

6.  A  lady  bought  a  dress  for  9  dollars  17  cents,  trimmings 
for  87^  cents,  a  paper  of  pins  for  6}  cents,  some  tape  for  4 
cents,  some  thread  for  8  cents,  and  a  comb  for  11  cent:; 
what  did  she  pay  for  all?  Ans.  $10.3375. 

7.  Paid  for  building  a  house  12175.75,  for  painting  the 
same  $240.3 7|-,  for  furniture  $605.40,  for  carpets  $140. 12^; 
what  was  the  cost  of  the  house  and  furnishing  ? 

8.  Bought  a  ton  of  coal  for  $6.08,  a  barrel  of  sugar  for 
$26,625,  a  box  of  tea  for  $16,  and  a  barrel  of  flour  for  $7.40 ; 
what  was  the  cost  of  all  ? 

9.  A  merchant  bought  goods  to  the  amount  of  $7425.50 ; 
he  paid  for  duties  on  the  same  $253.96,  and  for  freight 
$170.09 ;  what  was  the  entire  cost  of  the  goods  ? 

10.  I  bought  a  hat  for  $3.62^,  a  pair  of  shoes  for  $1},  an 
umbrella  for  $lf ,  a  pair  of  gloves  for  $.62J,  and  a  cane  for 
$.87 J;  what  was  the  cost  of  all  my  purchases  ?   Ans.  $8.25. 

SUBTRAOTIOlSr. 

164.  1.  A  man,  having  $327.50,  paid  out  $186.75  for  a 
horse  ;  how  much  had  he  left  ? 

OPERATION.  ANAI.YSIS.    Writing  the  less  number  un- 

$327.50  der  the  greater,  dollars  under  dollars,  cents 

186.75  under    cents,   etc.,  subtract  and    point  oflf 

in    the   result    as  ia  subtraction  of   deci- 

Ans.  $140.75  ^^j^ 

Rule.  I.  Write  the  subtrahend  under  the  minuend,  dollars 
under  dollars,  cents  under  cents. 

11.  SuUract  as  in  simple  numbers,  and  place  the  point  in 
the  remainder,  as  in  subtraction  of  decimals. 

Explain  the  process  of  subtraction.    Give  rule,  first  step.     Second. 


136  decimal  currency. 

Examples  for  Practice. 

2.  From  365  dollars  5  mills  take  267  dollars  1  cent  8 
mills.  Ans.  $97,987. 

3.  From  50  dollars  take  50  cents.  Ans.  $49.50. 

4.  From  100  dollars  take  1  mill  Ans.  $99,999. 

5.  From  1000  dollars  take  3  cents  7  mills. 

6.  A  man  bought  a  farm  for  $1575.24,  and  sold  it  foi 
$1834.16  ;  what  did  he  gain?  Ans.  $258.92. 

7.  Sold  a  horse  for  145  dollars  27  cents,  which  is  37  dol- 
lars 69  cents  more  than  he  cost  me  ;  what  did  he  cost  me  ? 

8.  A  merchant  bought  flour  for  $5.62^  a  barrel,  and  sold 
it  for  $6.84  a  barrel ;  what  did  he  gain  on  a  barrel  ? 

9.  A  gentleman,  having  $14725,  gave  $3560  for  a  store, 
and  $7015.87^  for  goods ;  how  much  money  had  he  left  ? 

10.  A  lady  bought  a  silk  dress  for  $13 J,  a  bonnet  for  $5 J, 
a  pair  of  gaiters  for  $1-|,  and  a  fan  for  $J  ;  she  paid  to  the 
shopkeeper  a  twenty  dollar  bill  and  a  five  dollar  bill ;  liow 
much  change  should  he  return  to  her  ?  Ans.  $3.75. 

Reduce  the  fractions  of  a  dollar  to  cents  and  mills.  « 

11.  A  gentleman  bought  a  pair  of  horses  for  $480,  a  har- 
ness for  $80.50,  and  a  carriage  for  $200  less  than  he  paid 
for  both  horses  and  harness ;  what  was  the  cost  of  the  car- 
riage? Ans.$3m.50. 

MULTIPLICATIOlSr. 

165.  1.  If  a  barrel  of  flour  cost  $6,375,  what  will  85 
barrels  cost? 

OPERATION. 

$6,375  Analysis.    Multiply  as  in  simple  num 

85  bers,  always  regarding  the  multiplier  as  an 

313'j'5  abstract  number,  and  point  off  from  the  right 

51000  hand  of  the  result,  as  in  multiplication  of 

^W5.  $541:875  decimals. 

^ 1 

Give  analysis  for  multiplication  in  decimal  currency. 


DIVISION.  137 

Rule.  Multiply  as  in  simple  numbers,  and  place  the  point 
in  the  product,  as  in  multiplication  of  decimals. 

Examples  for  Peactice. 

2.  If  a  cord  of  wood  be  worth  $4,275,  what  will  300  cords 
be  worth?  ^7Z5.  $1282.50. 

3.  What  will  175  barrels  of  apples  cost,  at  $2.45  per  bar- 
rel? Ans.  $428.75. 

4.  What  will  800  barrels  of  salt  cost,  at  $1.28  per  barrel  ? 

5.  A  grocer  bought  372  pounds  of  cheese  at  $.15  a  pound, 
434  pounds  of  coffee  at  $.12^  a  pound,  and  16  bushels  of 
potatoes  at  $.33  a  bushel ;  what  did  the  whole  cost  ? 

6.  A  boy,  being  sent  to  purchase  groceries,  bought  3 
pounds  of  tea  at  56  cents  a  pound,  15  pounds  of  rice  at  7 
cents  a  pound,  27  pounds  of  sugar  at  8  cents  a  pound ;  he 
gave  the  grocer  5  dollars ;  how  much  change  ought  he  to 
receive  ? 

7.  A  farmer  sold  125  bushels  of  oats  at  $.37^  a  bushel, 
and  received  in  payment  75  pounds  of  sugar  at  $.09  a  pound, 
12  pounds  of  tea  at  $.60  a  pound,  and  the  remainder  in 
cash ;  how  much  cash  did  he  receive  ?  Ans.  $32. 92 J. 

8.  A  man  bought  150  acres  of  land  for  $3975 ;  he  after- 
ward sold  80  acres  of  it  at  $32.50  an  acre,  and  the  remainder 
at  $34.25  an  acre  ;  what  did  he  gain  by  the  transaction? 

Ans.  $1022.50. 

DIVISION. 
166.  1.  If  125  barrels  of  flour  cost  $850,  how  much 
will  1  barrel  cost  ? 

OPERATION. 

125  )  $850.00  (  $6.80,  Ans.  Analysts.     Divide  as  in  sim- 

750  pie  numbers,  and  as  there  is  a 

^  remainder  after  dividing  the  dol- 

lars, reduce  the  dividend  to  cents, 
1000  by  annexing  two  ciphers,  and  coiif 

0  tinue  the  division. 

Rule.    Give  rule  for  division  in  decimal  currency. 


138  DECIMAL    CURRENCY. 

EuLE.  Divide  as  in  simple  numhers,  and  place  the  point 
in  the  quotient,  as  in  division  of  decimals. 

1.  In  business  transactions  it  is  never  necessary  to  carry  the  division  further  than 
to  mills  in  the  quotient. 

2.  If  the  dividend  will  not  contain  the  divisor  an  exact  number  of  times,  ciphers 
may  be  annexed,  and  the  division  continued  as  in  division  of  decimals.  In  this  case 
it  is  always  safe  to  reduce  the  dividend  to  mills,  or  to  3  more  decimal  places  thffn 
the  divisor  contains^  before  commencing  the  division. 

Examples  for  Practice. 

3.  If  33  gallons  of  oil  cost  $41.25,  what  is  the  cost  per 
gallon  ?  Ans,  11.25. 

3.  If  27  yards  of  broadcloth  cost  $94.50,  what  will  1  yard 
cost? 

4.  K  64  gallons  of  wine  cost  $136,  what  wiU  1  gallon 
cost?  ^»s.  $2,125. 

5.  At  12  cents  apiece,  how  many  pine-apples  can  be  bought 
for  $1.32  ?  Ans.  11. 

6.  If  1  pound  of  tea  costs  54  cents,  how  many  pounds 
can  be  bought  for  $405  ? 

7.  If  a  man  earns  $180  in  a  year,  how  much  does  he  earn 
a  month  ? 

8.  If  100  acres  of  land  cost  $2847.50,  what  will  1  acre 
cost?  Ans,  $28,475. 

9.  What  cost  1  pound  of  beef,  if  894  pounds  cost  $80.46  ? 

Ans,  $.09. 

10.  A  farmer  sells  120  bushels  of  wheat  at  $1.12^  a  bushel, 
for  which  he  receives  27  barrels  of  flour ;  what  does  the  flour 
cost  him  a  barrel  ? 

11.  A  man  bought  4  yards  of  cloth  at  $3.20  a  yard,  and 
37  pounds  of  sugar  at  $.08  a  pound  ;  he  paid  $6.80  in  cash, 
and  the  remainder  in  butter  at  $.16  a  pound ;  how  many 
pounds  of  butter  did  it  take  ?  Ans.  b^  pounds. 

12.  A  man  bought  an  equal  number  of  calves  and  sheep, 
paying  $166.75  for  them;  for  the  calves  he  paid  $4.50  a 
head,  and  for  the  sheep  $2. 75  a  head ;  how  many  did  he 
buy  of  each  kind  ?  Ans,  23. 


APPLICATIONS.  139 

13.  If  154  pounds  of  sugar  cost  $18.48,  what  will  1  pound 
cost  ? 

14.  A  merchant  bought  14  boxes  of  tea  for  $560  ;  it  being 
damaged  he  was  obliged  to  lose  $106.75  on  the  cost  of  it ; 
how  much  did  he  receive  a  box  ?  Ans,  $32.3  7^. 

Additional  Applications. 
Case  I. 

167.  To  find  the  cost  of  any  number  or  quantity, 
when  the  price  of  a  unit  is  an  aliquot  p2|-rt  of  one 
dollar. 

168.  An  Aliquot  Part  of  a  number  is  such  a  part  as 
will  exactly  divide  that  number ;  thus,  3,  5,  and  7i  are 
aliquot  parts  of  15. 

An  cUiqvot  part  may  be  a  whole  or  a  mixed  number,  while  a  factor  most  be  a 
whole  number. 

Aliquot  Pakts  op  Oke  Dollak. 


50    cents  =  |-  of  $1. 

33|  cents  =  i  ot  $1. 

25    cents  =  J  of  $1. 

20    cents  =  i  of  $1. 

16f  cents  =  i  of  $1. 


12J  cents  =  i  of  $1 

10    cents  =  ^  of  $1 

8^  cents  =  ^  of  $1 

6|  cents  =  ^  of  $1 

5    cents  =  -^^  of  $1 


1.  What  will  be  the  cost  of  3784  yards  of  flannel,  at  25 
cents  a  yard  ? 

OPEKATiON.  Analysis.     If  the  price  were  $1  a  yard, 

4  )  3784  the  cost  would  be  as  many  dollars  as  there  are 

J        *Qj.«  yards.    But  since  the  price  is  ^  of  a  dollar  a 

^  yard,  the  whole  cost  will  be  ^  as  many  dollars 

as  there  are  yards ;  or,  |  of  $3784  =  $3784  ^  4  ==  $946. 

KuLE.  Take  such  a  fractional  part  of  the  given  number  as 
the  price  is  part  of  one  dollar. 

Examples  foe  Practice. 

2.  What  cost  963  bushels  of  oats,  at  33  J  cents  per  bushel  ? 

Ans.  $321. 

Case  I  is  what  ?    What  is  an  aliquot  part  of  a  dollar  ?    Give  expla- 
nation.   Rule. 


140  DECIMAL    CURRENCY. 

3.  AYhat  cost  478  yards  of  delaine,  at  50  cents  per  yard  ? 

4.  What  cost  4266  yards  of  sheeting,  at  8^  cents  a  yard  ? 

A71S.  1355.50. 

5.  What  cost  1250  bushels  of  apples,  at  12|-  cents  per 
bushel?  Ans.  $156.25. 

6.  What  cost  3126  spools  of  thread,  at  6J  cents  per  spool  ? 

Ans.  $195,375. 

7.  At  16f  cents  per  dozen,  what  cost  1935  dozen  of  eggsr 

A71S.  $322.50. 

8.  Wliat  cost  56480  yards  of  calico,  at  12|-  cents  per  yard  ? 

9.  At  20  cents  each  what  will  be  the  cost  of  1275  salt 
barrels?  Ans.  $255. 

Case  II. 

169.   The  price  of  one  and  the  quantity  being 
given,  to  find  the  cost. 

1.  How  much  will  9  barrels  of  flour  cost,  at  $6.25  per 
barrel  ? 

OPERATION.  Analysis.    Since  one  barrel  cost  $6.25,  9 

$6.25  barrels  will  cost  9  times  $6.25,  and  $6.25  x 

9  9  =  $56.25. 

Ans.  $56.25 

Rule.  Multiply  the  price  of  one  hy  tlie  quantity. 

Examples  foe  Practice. 

2.  If  a  pound  of  beef  cost  9  cents,  what  will  864  pounds 
cost?  Ans.  $77.76. 

3.  What  cost  87  acres  of  goyemment  land,  at  $1.25  per 
acre? 

4.  What  cost  400  barrels  of  salt,  at  $1.45  per  barrel  ? 

Ans,  $580. 

5.  What  cost  10  chests  of  tea,  each  chest  containing  5^ 
pounds,  at  44  cents  per  pound  ? 

Case  II  is  what  ?    Give  explanation.    Rule. 


APPLICATIONS.  141 

Case  III. 

170.  The  cost  and  the  quantity  being  given,  to 
find  the  price  of  one. 

1.  If  30  bushels  of  corn  cost  120.70,  what  will  1  bushel 
cost? 

OPERATION.  Analysis.    If  30  bushels  cost  $20.70,  1 

3|0  )  $2|0.70  bushel  wiU  cost  ^  of  $30.70  ;  and  $20.70  h- 

^-  30  =  $.69. 

EuLE.  Divide  the  cost  ly  the  quantity. 

Examples  for  Peactice. 

2.  If  25  acres  of  land  cost  $175,  what  will  1  acre  cost  ? 

3.  If  48  yards  of  broadcloth  cost  $200,  what  will  1  yard 
cost?  Ans.  $4.16f. 

4.  If  96  tons  of  hay  cost  $1200,  what  will  1  ton  cost  ? 

5.  If  10  Unabridged  Dictionaries  cost  $56.25,  what  will  1 
cost?  ^?zs.  $5.62J. 

6.  Bought  18  pounds  of  tea  for  $11.70  ;  what  was  the 
price  per  pound  ?  Ans.  $.65. 

7.  If  53  pounds  of  butter  cost  $10.07,  what  will  1  pound 
cost? 

8.  A  merchant  bought  800  barrels  of  salt  for  $1016  ;  what 
did  it  cost  him  per  barrel  ? 

9.  If  343  sheep  cost  $874.65,  what  will  1  sheep  cost? 

Ans,  $2.55. 

10.  If  board  for  a  family  be  $684.37^  for  1  year,  how 
much  is  it  per  day  ?  Ans,  $1.87^. 

Case  IV. 

171.  The  price  of  one  and  the  cost  of  a  quantity 
being  given,  to  find  the  quantity. 

1.  At  $6  a  barrel  for  flour,  how  many  barrels  can  be 
bought  for  $840  ? 

Case  III  is  what  ?    Give  explanation.     Rule.     Case  IV  is  what  ? 


14^  DECIMAL    CUKREKCY. 

OPERATION.  ANAiiTSis.    Sincc  $6  will  buy  1  barrel 

6  )  840  of  flour,  $840  will  buy  ^  as  many  barrels 

as  there  are  dollars,  or  as  many  barrels  as 

Ans,  140  barrels.  ^q  ig  contained  times  in  $840;  840-^6 

=  140  barrels. 

Rule.  Divide  the  cost  of  the  quantity  ly  the  price  of  one. 

Examples  for  Practice. 

2.  How  many  dozen  of  eggs  can  be  bought  for  $5.55,  if 
one  dozen  cost  $.15  ?  Ans.  37  dozen. 

3.  At  $12  a  ton,  how  many  tons  of  hay  can  be  bought  for 
$216  ?  Ans.  18  tons. 

4.  How  many  bushels  of  wheat  can  be  bought  for  $2178.75, 
if  1  bushel  cost  $1.25?  Ans.  1743  bushels. 

5.  A  dairyman  expends  $643.50  in  buying  cows  at  $194- 
apiece  ;  how  many  cows  does  he  buy?         Ans.  33  cows. 

6.  At  $.45  per  gallon,  how  many  gallons  of  molasses  can 
be  bought  for  $52.65  ? 

7.  A  drover  bought  horses  at  $264  a  pair ;  how  many 
horses  did  he  buy  for  $6336  ? 

8;  At  $65  a  ton,  how  many  tons  of  railroad  iron  can  be 
bought  for  $117715  ?  Ans,  1811  tons. 

Case  V. 

172.  To  find  the  cost  of  articles  sold  by  the  100, 
or  1000. 

1.  What  cost  475  feet  of  timber,  at  $5.24  per  100  feet  ? 

FIRST  OPERATION. 

$5.24  Analysis.    If  the  price  were  |5.24  per 

475  foot,  the  cost  of  475  feet  would  be  475  x 

$5.24  =  $2489.    But  since  $5.24  is  the  price 

of  100  feet,  $2489  is  100  times  the  true 

value.     Therefore,  to  obtain  the  true  value, 

we  divide  $2489  by  100,  which  we  may  do 

100  )  $2489.00  by  cutting  off  two  figures  from  the  right, 

Ans    ^24  89  ^^^  *^®  result  is  $24.89.    Or, 

Give  explanation     Rule.    Case  V  is  what  ?    Give  first  explanation. 


2620 
3668 
2096 


APPLICATIONS.  143 

SECOND  OPEKATION.  ANALYSIS.     Since  1  foot  cost  Y^,  or  .01, 

$5.24  of  $5.24,  475  feet  will  cost  f^f,  or  4.75  times 

4.75  ^5.24,  which  is  $24.89. 

2620  For  the  same  reasons,  when  the  price  is  per  thxm- 

q/j/:.o  «awrf,  we  divide  the  product  by  1000,  or,  which  is  more 

convenient  in  practice,  we  reduce  the  given  quantity 

2096  to  thousands  and  decimals  of  a  thousand,  by  pointing 

^  off  three  figures  from  the  right  hand. 

$24.8900 

EuLE.  I.  Reduce  the  given  quantity  to  hundreds  and  deci- 
7nals  of  a  hundred,  or  to  thousands  and  decimals  of  a  thousand, 

II.  Multiply  the  price  hy  the  quantity,  and  point  off  in  the 
result  as  in  multiplication  of  decimals. 

The  letter  C  is  used  to  indicate  hundreds,  and  M  to  indicate  thousands. 

Examples  for  Practice. 

2.  What  will  42650  bricks  cost,  at  $4.50  per  M  ? 

Ans.  $191,925. 

3.  What  is  the  freight  on  2489  pounds  from  Boston  to 
New  York,  at  $.85  per  100  pounds  ?  Ans.  $21,156. 

4.  What  wiU  7842  feet  of  pine  boards  cost,  at  $17.25 
per  M  ?  Ans.  $135,274. 

5.  What  cost  2348  pine-apples,  at  $12^  per  1000  ? 

6.  A  broom  maker  bought  1728  broom-handles,  at  $3  per 
1000  ;  what  did  they  cost  him  ? 

7.  What  is  the  cost  of  2400  feet  of  boards,  at  $7  per  M  ; 
865  feet  of  scantling,  at  $5.40  per  M  ;  and  1256  feet  of  lath, 
at  $.80  per  0  ?  Ans.  $31,519. 

8.  What  will  be  the  cost  of  1476  pounds  of  beef,  at  $4.37i 
per  hundred  pounds  ? 

Case  VL 

173.  To  find  the  cost  of  articles  sold  by  the  ton 
of  2000  pounds. 

1.  How  much  will  2376  pounds  of  hay  cost,  at  $9.50  per 
ton? 

Give  second  explanation.    Rule,  first  step.    Second.    Case  VI  is  what  ? 


144  DECIMAL    CUEREKCY, 


OPERATION.  Analysis.    Since  1  ton,  or  2000  pounds,  cost 

2  )  $9.50  $9.50,  1000  pounds,  or  ^  ton,  will  cost  |-  of  $9.50, 

or  $9.50  H-  2  =  $4.75.  One  pound  will  cost 
■n^,  or  .001,  of  $4.75,  and  2376  pounds  will 
cost  fro,  or  2.376  times  $4.75,  which  is  $11,286. 


$4.75 
2.376 


$11.28600 

Rule.  I.  Divide  the  price  of  1  ton  ly  2,  and  the  quotient 
will  he  the  'price  of  1000  pounds. 

II.  Multiply  this  quotient  hy  the  given  number  of  pounds 
expressed  as  thousandths,  as  in  Case  V, 


Examples  for  Peactice. 

2.  At  17  a  ton,  what  will  1495  pounds  of  hay  cost  ? 

Ans.  $5.2325. 

3.  At  $8.75  a  ton,  what  cost  325  pounds  of  hay  ? 

Ans.  $1,421. 

4.  What  is  the  cost  of  3142  pounds  of  plaster,  at  $3.84 
per  ton?  '  Ans.  $6,032. 

5.  What  is  the  cost  of  1848  pounds  of  coal,  at  $5.60  per 
ton? 

6.  Bought  125  sacks  of  guano,  each  sack  containing  148 
pounds,  at  $18  a  ton ;  what  was  the  cost  ? 

7.  What  must  be  paid  for  transporting  31640  pounds  of 
railroad  iron  from  Philadelphia  to  Richmond,  at  $3.05  per 
ton?  Ans.  $48,251. 

BILLS. 

174.  A  Bill,  in  business  transactions,  is  a  written  state- 
ment of  articles  bought  or  sold,  together  with  the  prices  of 
each,  and  the  whole  cost. 

Find  the  cost  of  the  several  articles,  and  the  amount  or 
footing  of  the  following  bills. 

Give  explanation.  Rule.  What  is  a  bill  ?  Explain  the  manner  of 
making  out  a  bill. 


BILLS 


1^5 


ttlr,  JoHiq^  EiCE, 


(1.) 
New  Yoek,  June  20,  1869. 

BoH  of  Baldwin  &  Sherwood. 

7  yds.  Broadcloth,  @  83.60 

9    "    Satinet,         "     1.12^ 
12    "    Vesting,       "      .90 
24    "    Cassimere,    ''     1.37^ 
32    ''    Flannel,       ''       .65 


Redd  Payment, 


$99,925 
BALDWii?^  &  Sherwood. 


(2.) 


BosT02^r,  Jan.  1,  1870. 


Daniel  Chapmak  &  Co., 

BoH  of  Palmer  &  Brother. 
67  pairs  Calf  Boots,  @  $3.75 


108 
75 
27 
35 
50 


''    Thick   ' 
''    Gaiters, 


Buskins, 
Slippers, 
Rubbers, 


2.62 

1.12 

.86 

.70 

1.04 


Re<fd 

Payment, 

$717.93 
Palmer  &  Brother, 

By  Geo.  Baker. 

(3.) 

Charlestoi^,  Sept.  6,  1875. 
G.  B.  Grai^nis, 

BoH  of  Stewart  &  Hammokd. 

325  lbs.  A.  Sugar, 

@ 

$.07 

148 

"    B.      " 

it 

.06i 

286 

"    Eice, 

(I 

.05 

95 

''    0.  J.  Coffee, '' 

.12i 

60  boxes  Oranges, 

(I 

2.75 

75 

**     Lemons, 

a 

3.62i 

12 

*'    Raisins, 

a 

2.85 

Eec^d  Payment,  by  note  at  4  mo.        ^^01.75 
r.p.  7  Stewart  &  Hammond. 


J-46  DECIMAL    CURRENCY. 

(4.) 

St.  Louis,  Oct.  15,  1878. 
Messrs.  Osborn  &  Eaton^, 

BoH  of  Rob't  H.  Carter  &  Gg^ 

20000  feet  Pine  Boards,  @  $15.00  per  M. 

7500    ''    Plank,  "      9.50    '' 

10750    ''    Scantling,      ''      6.25    " 

3960    ''   Timber,  "      2.62^  " 

5287    "        "  "      3.00    '' 


Bec^d  Payment, 

$464.6935 

Rob't  H.  Carter  &  Co. 

(5.) 

CiN^ciirif  ATI.  Mav  3, 1871. 

Mr.  J.  C.  Smith, 

BoH  of  Silas  Johnson, 

25  lbs. 

Coffee  Sugar,           @ 

$.11 

5    '' 

Y.  H.  Tea, 

.62^ 

26    '' 

Mackerel,                  " 

.061 

4  gal. 

Molasses,                   " 

.42 

46  yds. 

Sheeting,                   '^ 

.09 

30   " 

Bleached  Shirting,    '^ 

.14 

6  skeins 

3  Sewing  Silk,             *^ 

.04 

4  doz. 

Buttons,                    " 

.12 

Chgd.  in  %.  ^^^-^^ 

Silas  Johnson, 

Per  John  Wisk. 

Promiscuous  Examples. 

1.  What  will  62.75  tons  of  potash  cost,  at  $124.35  per  ton  ? 

Ans.  $7802.9625. 

2.  What  cost  15  pounds  of  butter,  at  $.17  a  pound? 

Ans.  $2.55. 

3.  A  cargo  of  com,  containing  2250  bushels,  was  sold  for 
$1406.25  ;  what  did  it  sell  for  per  bushel  ?  Ans,  $4- 


PROMISCUOUS    EXAMPLES.  147 

4.  If  12  yards  of  cloth  cost  $48.96,  what  will  one  yard 
cost? 

5.  A  trayeled  325  miles  by  railroad,  and  0  traveled  .45  of 
that  distance  ;  how  far  did  0  travel  ?    Ans.  146.25  miles. 

6.  If  36.5  bushels  of  corn  grow  on  one  acre,  how  many 
acres  will  produce  657  bushels  ?  Ans.  18  acres. 

7.  Bought  a  horse  for  $105,  a  yoke  of  oxen  for  $125,  4 
cows  at  $35  apiece,  and  sold  them  all  for  $400 ;  what  was 
gained  or  lost  in  the  transaction  ? 

8.  A  man  bought  28  tons  of  hay  at  $19  a  ton,  and  sold 
it  at  $15  a  ton  ;  what  did  he  lose  ?  Ans,  $112. 

9.  If  a  man  travel  4|  miles  an  hour,  in  how  many  hours 
can  he  travel  34|-  miles  ?  Ans.  7.5  hours. 

10.  At  $.31:J  per  bushel,  how  many  bushels  of  potatoes 
can  be  bought  for  $9  ?  Aiis.  28.8  bushels. 

11.  If  a  man's  income  be  $2000  a  year,  and  his  expenses 
$3.50  a  day,  what  will  he  save  at  the  end  of  a  year,  or  365 
days? 

12.  A  merchant  deposits  in  a  bank,  at  one  time,  $687.25, 
and  at  another,  $943.64 ;  if  he  draw  out  $875.29,  what  will 
remain  in  the  bank  ? 

13.  Bought  288  barrels  of  flour  for  $1728,  and  sold  one- 
half  the  quantity  for  the  same  price  I  gave  for  it,  and  the 
other  half  for  $8  per  barrel;  what  did  I  receive  for  the 
whole  ?  Ans.  $2016. 

14.  What  will  eight  hundred  seventy-five  thousandths  of 
a  cord  of  wood  cost,  at  $3.75  per  cord?      A^is.  $3,281  +  . 

15.  A  drover  bought  cattle  at  $46.56  per  head,  and  sold 
them  at  $65.42  per  head,  and  thereby  gained  $3526.82 ;  how 
many  cattle  did  he  buy  ?  Ans.  187. 

16.  If  36.48  yards  of  cloth  cost  $54.72,  what  wiU  14.25 
yards  cost  ?  Ans.  $21. 375. 

17.  A  house  cost  $3548,  which  is  4  times  as  much  as  the 
furniture  cost;  what  did  the  furniture  cost?     Ans.  $887. 

18.  How  many  bushels  of  onions  at  $.82  per  bushel,  can 
bo  bought  for  $112.34  ? 


148  DECIMAL    CURRENCY. 

19.  If  46  tons  of  iron  cost  13461.50,  what  will  5  tons  cost? 

20.  A  gentleman  left  his  widow  one-third  of  his  property, 
worth  124000,  and  the  remainder  was  to  be  divided  equally 
among  5  cliildren ;  what  was  the  portion  of  each  child  ? 

Ans.  $3200. 

21.  A  man  purchased  one  lot,  containing  160  acres  of 
land,  at  |>1.25  per  acre  ;  and  another  lot,  containing  80 
acres,  at  $5  per  acre  ;  he  sold  them  both  at  $2.50  per  acre; 
what  did  he  gain  or  lose  in  the  transaction  ? 

22.  A  druggist  bought  54  gallons  of  oil  for  $72.90,  and 
lost  6  gallons  of  it  by  leakage.  He  sold  the  remainder  at 
$1.70  per  gallon  ;  what  did  he  gain  ?  Ans.  $8.70. 

23.  A  miller  bought  122-|-  bushels  of  wheat  of  one  man, 
and  75 J  bushels  of  another,  at  $.93}  per  bushel.  He  sold 
60  bushels  at  a  profit  of  $12.50 ;  if  he  sell  the  remainder  at 
$.81 J  per  bushel,  what  will  be  his  entire  gain  or  loss? 

Ans.  $4,718+  loss. 

24.  A  laborer  receives  $1.40  per  day,  and  spends  $.75  for 
his  support ;  how  much  does  he  save  in  a  week  ? 

25.  How  many  pounds  of  butter,  at  $.16  per  pound,  must 
be  given  for  39  yards  of  sheeting,  at  $.08  a  yard  ? 

Ans.  19|-  pounds. 

26.  What  cost  23487  feet  of  hemlock  boards,  at  $4.50  per 
1000  feet  ?  Ans.  $105.6915. 

27.  A  man  has  an  income  of  $1200  a  year ;  how  much 
must  he  spend  per  day  to  use  it  all  ? 

28.  Bought  28  firkins  of  butter,  each  containing  56 
pounds,  at  $.17  per  pound;  what  was  the  whole  cost  ? 

29.  A  merchant  bought  16  bales  of  cotton  cloth,  each  bale 
containing  13  pieces,  and  each  piece  26  yards,  at  $.07  per 
yard ;  what  did  the  whole  cost  ?  Ans,  $378.56. 

30.  What  cost  4868  bricks,  at  $4.75  per  M  ? 

31.  A  farmer  sold  27  bushels  of  potatoes,  at  $.33|-  per 
bushel;  28  bushels  of  oats,  at  $.25  per  bushel ;  and  19  bush- 
els of  corn,  at  $.50  per  bushel ;  what  did  he  receive  for  the 
who'.e?  Ans.  $25.50. 


PROMISCUOUS    EXAMPLES.  149 

32.  John  runs  32  rods  in  a  minute,  and  Henry  pursues 
him  at  the  rate  of  44  rods  in  a  minute  ;  how  long  will  it 
take  Henry  to  overtake  John,  if  John  have  8  minutes  the 
start?  Ans.  211- mmntes, 

33.  If  4|  barrels  of  flour  cost  132.3,  what  will  7^  barrels 
cost  ?  Ans.  $51. 

34.  If  .875  of  a  ton  of  coal  cost  15.635,  what  will  9J  tons 
cost  ?  Ans.  $59.57. 

35.  For  the  first  three  years  of  business,  a  trader  gained 
$1200.25  a  year ;  for  the  next  three,  he  gained  $1800.62  a 
year,  and  for  the  next  two  he  lost  $950.87  a  year  ;  supposing 
his  capital  at  the  beginning  of  trade  to  have  been  $5000,  what 
was  he  worth  at  the  end  of  the  eighth  year?  Ans.  $12100.87. 

36.  What  will  be  the  cost  of  18640  feet  of  timber,  at  $4.50 
per  100?  Ans.  $Sd8.S0. 

24 

37.  Eeduce  ^  to  a  decimal  fraction.         Ans.  .78125. 

38.  What  will  1375  pounds  of  potash  cost,  at  $96.40  per 
ton  ?  Ans.  $66,275. 

39.  Eeduce  .5625  to  a  common  fraction.  Ans.  -^. 

40.  Eeduce  -^,  .62J,  .37^^^  f,  to  decimals,  and  find  their 
sum.  Ans.  1.464375. 

41.  A  man's  account  at  a  store  stands  thus  : 

Dr.  Cr. 

$4,745  $2.76^ 

2.62^  1.245 

1.27  .62J 

.45  3.45 

5.28J  1.87i 

What  is  due  the  merchant  ?  Ans.  $4.41|-. 

42.  A  gardener  sold,  from  his  garden,  120  bunches  of 
onions  at  $.12^  a  bunch,  18  bushels  of  potatoes  at  $.62|-  per 
bushel,  47  heads  of  cabbage  at  $.  07  a  head,  6  dozen  cucum- 
bers at  $.18  a  dozen  ;  he  expended  $1.50  in  spading,  $1.27 
for  fertilizers,  $1.87  for  seeds,  $2.30  in  planting  and  hoeing  ; 
what  were  the  profits  of  his  garden  ?  Ans.  $23.68. 


150 


REDUCTION. 


EEDUCTIOW. 

175.  A  Compound  Number  is  a  number  whose  value 
is  expressed  in  two  or  more  different  denominations. 

176.  Reduction  is  the  process  of  changing  a  number 
from  one  denomination  to  another  without  altering  its  value. 

Reduction  is  of  two  kinds,  Descending  and  Ascending. 

177.  Reduction  Descending  is  changing  a  number  of 
one  denomination  to  another  denomination  of  less  unit 
value;  thus,  $1=10  dimes =100  cents =1000  mills. 

178.  Reduction  Ascending  is  changing  a  number  of  one 
denomination  to  another  denomination  ot  greater  unit  valtte; 
thus,  1000  mills=100  cents=10  dimes=$l. 

179.  A  Scale  is  a  series  of  units,  increasing  or  decreas- 
.ing,  according  to  a  certain  law. 

CURRENCY. 

180.  I.  UiiriTED  States  Monet. 

Table. 

10  Mills  (m.)  make  1  Cent ct.  f  10000  m. 

10  Cents  "     IDime d.  ^„_J    1000  ct. 

10  Dimes  "     1  Dollar $.  "^  ^—  1      100  d. 

10  Dollars  "     1  Eagle E.  [       10 1. 

Unit  Equivalents. 
ct. 
d.  1  = 

$  1  =     10  = 

E.       1  =    10  =    100  = 
1  =  10  =  100  =  1000  =  10000 
Scale— uniformly  10. 

Canada  Money. 
The  currency  of  the  Dominion  of  Canada  is  decimal,  and 
the  table  and  denominations  are  the  same  as  those  of  the 
United  States  money. 

The  currency  of  the  whole  Dominion  of  Canada  was  made  uniform  July  1, 1871. 
Before  the  adoption  of  the  decimal  eystem,  pounds,  shillingP,  and  pence  were  used. 

The  Silver  Coins  are  the  50-cent  piece,  25-cent  piece,  10-cent  piece, 
and  5-cent  piece.  The  20-cent  piece  is  no  longer  coined.  The  Bronze 
Coin  is  the  cent. 

The  Odd  Coin  used  in  Canada  is  the  British  Sovereign,  worth  |4.8C|, 
and  the  Half- Sovereign. 


m. 

10 

100 

1000 


compoukd   numbers.  151 

11.    English  Monet. 
181.  English  Currency  is  the  currency  of  Great  Britain, 

Table. 

U.  S.  Value. 

4  Farthings  (far.)  make  1  Penny d $0.0202  + 

12  Pence  "     1  Shilling s 2433  + 

20  ShiUings  *  *     1  Pound  or  Sov £.,  or  sov. . . . .  $4.8665 

Unit  Equivalents. 

d.       far. 

8.         1  =     4 

£.        1  =    12  =    48 

1  =  20  =  240  =  960 

Scale  ascending,  4,  12,  20 ;  descending,  20,  12,  4. 

1.  Farthings  are  generally  expressed  as  fractions  of  a  penny ;  thus,  1  far.,  some* 
times  called  1  quarter  (qr.),  =  |d. ;  3  fer.  =  |d. 

2.  The  gold  coins  are  the  sovereign  (  =  £1),  and  the  half-sovereign. 

3.  The  silver  coin^  are  the  crown  ( =  58.),  the  half-crown  ( =  2s.  6d.),  the  florin,  the 
ehilling,  and  the  sixpenny,  fourpenny,  and  threepenny  pieces. 

4.  The  copper  coins  are  the  penny,  halfpenny,  and  farthing, 

5.  The  guinea  ( =  2l8.)  and  the  half-guinea  ( =  10s.  6d.  sterling),  are  old  gold  coins, 
and  are  no  longer  coined. 

6.  In  France  accounts  are  kept  in  francs  and  decimes.    A  franc  is  equal  to  $.198 
U.  S.  money. 

Case  I. 

183.  To  perform  reduction  descending. 

1.  Keduce  £21 18s.  lOd.  2  far.  to  farthings. 

operation.  Analysis.    Since  in  £1  there 

£21  IBs.  lOd.  2  far.  are  20s.,  in  £21  there  are  20s.  x 

20  21  =  420s.,  and  IBs.  in  the  given 

number  added,  makes  438s.  in  £21 

18s.     Since  in  Is.  there  are  12d., 

in  438s.  there  are   12d.  x  438  = 

6266d.  5256d.,  and  lOd.  in  the  given  num- 

4  ber  added,  makes  5266d.  in  £21 

18s.  lOd.    Since  in  Id.  there  are 
21066  far.,  Ans.  4  f^r.,  in  5266d.  there  are  4  far.  x 

=  21064  far.,  and  2  far.  in  the  given  number  added,  makes  21066 
far.  in  £21  18s.  lOd.  2  far. 


438s. 
12 


152  REDUCTlOiT. 

Rule.  I.  Multiply  the  highest  dcnominatmi  of  the  given 
number  hy  that  number  of  the  scale  which  will  reduce  it  to 
the  next  loioer  denomination,  adding  to  the  product  the  given 
number,  if  any,  of  that  loioer  denominatioti. 

II.  Proceed  in  the  same  manner  with  the  results  obtained 
in  each  lower  denomination,  until  the  reduction  is  brought  to 
the  denomination  required. 

Case  IL 

183.  To  perform  rediiction  ascending. 

1.  Reduce  21066  farthings  to  pounds. 

OPERATION.  Analysis.    First  divide  21066 

4  )  21066  far.  far.  by  4,  since  there   are  \  as 

1  o  \  "nnnj    ,     o  £  many  pence  as  farthings,  and  we 

12  )_o266d.  +  2  fax.  ^^^  ,^^,  ^^^^  ^^^  ^  52^^^  ^  ^ 

2|0  )  43|8s.  +  lOd.  remainder  of  2  far.    Next  divide 

r>2i    I  1  Qq  5266d.  by  12,  since  there  are  ^^ 

J         rt^^^^\./s-irt<.  ^  many  shillings  as  pence,  and 

Ans.  £n  18s.  lOd.  3  far.        ^,  ^  J  ,^,,  ^^^  I  ^^  ^ 

lOd.  Lastly  divide  438s.  by  20,  since  there  are  ^V  ^  many  pounds 
as  shillings,  and  we  find  that  438s.  =  £12  +  18s.  The  last  quotient 
with  the  several  remainders  annexed  in  the  order  of  the  succeeding 
denominations,  gives  the  answer  £21  18s.  lOd.  2  far. 

Rule.  I.  Divide  the  given  number  by  that  number  of  the 
scale  which  will  reduce  it  to  the  next  higher  deno7nination. 

IL  Divide  the  quotient  by  the  next  higher  number  in  the 
scale  ;  and  so  proceed  to  the  highest  denomination  required. 
The  last  quotient,  with  the  several  remainders  annexed  in  a 
reversed  order,  will  be  the  answer. 

Reduction  descending  and  redaction  ascending  mntaally  prove  eacSi  other. 

Examples  for  Practice. 

1.  In  14194  farthings  how  many  pounds  ? 

2.  In  £14  los.  8d.  2  far.  how  many  farthings  ? 

3.  In  15359  farthings  how  many  pounds? 

4.  In  46  sov.  12s.  2d.  how  many  pence  ? 

5.  In  11186  pence  how  many  sovereigns? 


COMPOUND    KUMBEES, 


153 


WEIGHTS. 

184.  Weight  is  the  measure  of  gravity,  and  varies  ac- 
cording to  the  quantity  of  matter  a  body  contains.  Three 
scales  of  weight  are  used  in  the  United  States  and  Great 
Britain,  namely,  Troy,  Apothecaries',  and  Avoirdupois. 

I.    Teoy  Weight. 

185.  Troy  Weight  is  used  in  weighing  gold,  silver,  and 
jewels,  and  in  philosophical  experiments. 

Table. 

24  Grains  (gr.)  make  1  Pennyweight pwt. 

20  Pennyweights  "       1  Ounce oz. 

12  Ounces  "       1  Pound lb. 

Unit  Equivalents. 

pwt.       gr. 

oz.         1  =     24 

lb.       1  =    20  r=    480 

1  =  12  =  240  =  5760 

SOALE— ascending,  24,  20,  12  ;  descending,  12,  20,  24. 

Examples  toe  Pkactice. 


1.    How  many  grains  in 
141b.  10  oz.  18  pwt.  22  gr.? 

OPERATION. 

14  lb.  10  OZ.  18  pwt.  22  gr. 
12 

178  OZ. 
20 


3578  pwt. ; 
3578  pwt.  X  24  =  85894  gr., 
Ans. 


2.  How  many  pounds  in 
85894  grains  ? 

OPERATION. 

24 )  85894  gr. 
20  )  3578  pwt.  +  22  gr. 
12  )178  oz.  +  18  pwt. 
14  lb.  +  10  oz. 


Ans.   14  1b.  10  oz.  18  pwt 
22  gr. 

3.  In  5  lb.  7  oz.  12  pwt.  9  gr.,  how  many  grains  ? 

4.  In  32457  grains  how  many  pounds  ? 


Define  weight.    Troy  weight.    Repeat  the  table.     Give  the  scale. 


154 


R:feDUCTION 


5.  Eeduce  41760  grains  to  pounds.         Ans.  7  lb.  3  oz. 

6.  A  miner  had  14  lb.  10  oz.  18  pwt.  of  gold  dust ;  what 
was  it  worth  at  $.75  a  pwt.  ?  Ans.  $2683.50. 

7.  How  many  spoons,  each  weighing  2  oz.  15  pwt.,  can  be 
made  from  5  lb.  6  oz.  of  silver  ?  Ans,  24. 

8.  A  goldsmith  manufactured  1  lb.  1  pwt.  16  grs.  of  gold 
into  rings,  each  weighing  4  pwt.  20  gr. ;  he  sold  the  rings  for 
$1.25  apiece ;  what  did  he  receive  for  them?    Ans,  $62.50. 

II.    Apothecaeies'  Weight. 
186.  Apothecaries'  Weight  is  used  by  Apothecaries 
and  physicians  in  compounding  dry  medicines  ;  but  medi- 
cines are  bought  and  sold  by  Avoirdupois  Weight. 

Table. 
20  Grains  (gr.  xx)  make  1  Scruple sc.  or  ^. 

3  Scruples  (3 iij)       **      1  Dram dr.  or  3. 

8  Drams  ( 3  viij)        '*      1  Ounce oz.  or  ^ . 

12  Ounces  (  §  xij)        "      1  Pound lb.  or  a. 

Unit  EQurvALENTS. 


dr. 
>z.       1  = 

1=8  = 


sc. 
1  = 
3  = 

24  = 


20 
60 

480 


Tb 

1  =  13  =  96  =  288  =  5760 
Scale — ascending,  20,  3,  8,  12 ;  descending,  12,  8,  3,  20. 

Examples  for  Practice. 


1.  How  many  gr.  in  12  ft) 
8?  33  13  15  gr. 

OPERATION. 

12ft)8!  3  3  I3l5gr.? 
12 


152  § 
8 


1219  3  ; 


1219  3  X  3=36583  ; 
36583  X  20=73175  gr.,^?i5. 


gr. 


2.  How  many  ft)  in  7317?* 

OPERATION. 

2|0 )  7317|5  gr. 
3)  36583 +15  gr. 
8)1219  3+13 
12)  152  g  +3  3 
12ft)  +  8? 

Ans.  12  ft)  8?  3  3  l3  15 gr. 


Define  apothecaries'  weight.    Repeat  the  table.    Give  the  scale. 


OMPOUND    NUMBERS.  155 

3.  In  16  lb.  11  oz.  7  dr.  2  sc.  19  gr.,  how  many  grains  ? 

4.  Eeduce  47ft)  6  §  4  3  to  scruples.        Ans.  13692  sc. 

5.  How  many  pounds  of  medicine  would  a  physician  use 
in  one  year,  or  365  days,  if  he  averaged  daily  5  prescriptions 
of  20  grains  each  ?  Ans.  6ft).  4  |  13. 

Til.    Avoirdupois  Weight. 
187.  Avoirdupois  Weight  is  used  for  all  the  ordinary 
purposes  of  weighing. 

Table. 

16  Ounces  make  1  Pound lb. 

100  Pounds  "       1  Hundred- weight cwt. 

20  Cwt.,  or  2000  lbs.,  "      1  Ton T. 

Unit  Equivalents. 
lb.  oz. 

cwt.         1  =       16 
T.        1  =    100  =    1600 
1  =  20  =  2000  =  32000 
Scale— ascending,  16,  100,  20 ;  descending,  20, 100,  16. 
The  long  or  gross  ton,  hundred-weight,  and  quarter  were  formerly  in  common 
use  ;  but  they  are  now  seldom  used  except  in  estimating  English  goods  at  the  U.  S. 
custom-houses,  and  in  freighting  and  wholesaling  coal  at  the  mines. 

Long  Ton  Table. 

16  Ounces  make  1  Pound,  marked    lb. 

28  Pounds  "       1  Quarter,  "         qr. 

4  Quarters  "      1  Hundred- weight,      "       cwt. 

20  Cwt.  =  2240  lb.    «      1  Ton,  "  T. 

The  following  denominations  are  also  in  use  : 

100  Pounds  of  Grain  or  Flour  make  1  Cental. 

100       "  Dry  Fish  "      1  Quintal, 

100       "  Nails  "      1  Keg. 

196        "  Flour  .    "      1  Barrel. 

200        "  Pork  or  Beef  "      1  Barrel. 

280        "  Salt  at  N.  Y.  S.  works    "      1  Barrel. 

56        "  "  "  "  **      1  Bushel. 

240        "  Lime  "      1  Cask. 

32        ♦'  Oats  "      1  Bushel. 

56        "  Com  "      1  Bushel. 

60        "  Wheat  "      1  Bushel. 

Define  avoirdupois  weight.  Repeat  the  table.  Give  the  scale.  The 
long  ton  table.    What  other  denominations  are  in  use  ? 


156  REDUCTION. 


Examples  for  Practice. 
1.  In  25  T.  15  cwt.  70  lb, 
how  many  pounds  ? 

OPERATION. 

25  T.  15  cwt.  70  lb. 
_20 

515  cwt. 
100 


2.   In  51570  pounds  how 
many  tons  ? 

OPERATION. 

100 )  51570  lb. 
2|0)51|5cwt.  +  70  1b. 
25  T.  +  15  cwt. 
Ans.  25  T.  15  cwt.  70  lb. 


51570  lb.,  Ans. 

3.  Reduce  3  T.  14  cwt.  74  lb.  12  oz.  to  ounces. 

4.  Eeduce  119596  ounces  to  tons. 

5.  A  tobacconist  bought  3  T.  15  cwt.  20  lb.  of  tobacco,  at 
22  cents  a  pound  ;  what  did  it  cost  him  ?      Ans.  $1654.40. 

6.  What  will  115  pounds  of  hay  cost,  at  $10  per  ton? 

7.  A  grocer  bought  10  barrels  of  sugar,  each  weighing 

2  cwt.  17  lb.,  at  6  cents  a  pound  ;  5  barrels,  each  weighing 

3  cwt.  6  lb.,  at  7-J  cents  a  pound  ;  he  sold  the  whole  at  an 
average  price  of  8  cents  a  pound  ;  what  was  his  whole  gain? 

Ans.  $51.05. 

8.  Paid  $360  for  2  tons  of  cheese,  and  retailed  it  for  12 J 
cents  a  pound ;  what  was  my  whole  gain  ?       A  ns.  $140. 

9.  If  a  person  buy  10  T.  6  cwt.  3  qr.  14  lb.  of  Enghsh  iron, 
by  the  long  ton  weight,  at  6  cents  a  pound,  and  sell  the  same 
at  $130  per  short  ton,  what  will  he  gain  ?     Ans.  $115.85. 

10.  A  farmer  sold  2  loads  of  corn,  weighing  2352  lbs.  each, 
at  $.90  per  bu. ;  what  did  he  receive  ?  A^is.  $75.60. 

11.  How  many  pounds  in  300  barrels  of  flour  ? 

12.  A  grocer  bought  3  barrels  of  salt  at  $1.25  j^er  barrel, 
and  retailed  it  at  |  of  a  cent  per  pound  ;  what  did  he  gain  ? 

Atis.  $2.55. 
Standard  of  Weight. 
188.  In  the  year  1834  the  U.  S.  government  adopted  a 
uniform  standard  of  weights  and  measures,  for  the  use  of  the 
custom-houses,  and  the  other  branches  of  business  connected 
with  the  general  government. 


COMPOUND    NUMBERS.  157 

189.  The  United  States  standard  unit  of  weight  is  the 
Troy  pound  of  the  mint,  which  is  the  same  as  the  imperial 
standard  pound  of  Great  Britain,  and  is  determined  as  fol- 
lows :  A  cubic  inch  of  distilled  water  in  a  vacuum,  Aveighed 
by  brass  weights,  also  in  a  vacuum,  at  a  temperature  of  62° 
Fahrenheit's  thermometer,  is  equal  to  252.458  grains,  of 
which  the  standard  Troy  pound  contains  5760. 

190.  Tlie  U.  8.  Avoirdupois  pound  is  determined  from 
the  standard  Troy  pound,  and  contains  7000  Troy  grains. 
Hence,  the  Troy  pound  is  4ofo  =  ifi  of  an  avoirdupois 
pound.  But  the  Troy  ounce  contains  -S^f^  :=  480  grains, 
and  the  avoirdupois  ounce  -2^^  =  437.5  grains ;  and  an  ounce 
Troy  is  480—437.5=42.5  grains  greater  than  an  ounce  avoir- 
dupois. The  pound,  ounce,  and  grain.  Apothecaries'  weight, 
are  the  same  as  the  like  denominations  in  Troy  weight,  the  only 
diif  erence  in  the  two  tables  being  in  the  divisions  of  the  ounce. 

191.  COMPAEATIVE  TABLE  OF  WEIGHTS. 

Troy.  Apothecaries'.         Avoirdupois. 

1  pound  =  5760  grains,   =  5760  grains,    =  7000  grains. 
1  ounce  =    480      "        =480       "        =  437.5       " 
175  pounds,  =    175  pounds,  =  144  pounds. 

Examples  foe  Practice. 

1.  An  apothecary  bought  5  lb.  10  oz.  of  rhubarb,  by 
avoirdupois  weight,  at  50  cents  an  ounce,  and  retailed  it  at 
12  cents  a  dram  apothecaries'  weight ;  what  did  he  gain  ? 

2.  Change  424  drams  apothecaries'  weight  to  Troy  weight. 

3.  Change  20  lb.  8  oz.  12  pwt.  Troy  weight  to  avoirdu- 
pois weight.  Ans.  17-^  lb. 

4.  Bought  by  avoirdupois  weight  20  lb.  of  opium,  at  40 
cents  an  ounce,  and  sold  the  same  by  Troy  weight  at  50 
cents  an  ounce  ;  what  did  he  gain  ?  Ans.  $17.83 J. 

What  is  the  U.  S.  standard  of  weight  ?  How  obtained  ?  How  is 
the  avoirdupois  pound  determined  ?  How  is  the  apothecaries'  pound 
determined?  What  are  the  values  of  the  denominations  of  Troy, 
avoirdupois,  and  apothecaries*  weight  ? 


158  EEDUCTION. 

MEASUEES    OF    EXTENSION. 

192.  Extension  has  one  or  more  of  the  dimensions — 
length,  breadth,  and  thickness. 

A  Line  has  only  one  dimension — length, 

A  Surface  or  Area  has  two  dimensions — length  and 
Ireadth, 

A  Solid  or  Body  has  three  dimensions — length,  breadth, 
and  thickness. 

I.  Long  Measure. 

193.  Long  Measure,  also  called  Linear  Measure,  is 
used  in  measuring  lines  and  distances. 

Table. 

13    Indies  (in.)        make  1  Foot ft. 

3    Feet  "      1  Yard yd. 

5i  Yd.,  or  16^  ft.,       "      1  Rod rd. 

320    Rods  ty-  yy^t^^H**     1  Statute  Mile ...  .mi 

Unit  Equivalents. 

ft.  in. 

yd.  1    =       12 

rd.  1    =       3    =       36 

mi.        1  =       51  =     16i  =     198 

1  =  320  =  1760    =  5280    =  63360 

Scale— ascending,  12,  3,  ^,  320  ;  descending,  320,  5^,  3,  13. 

The  following  denominations  are  also  in  use : 

3  Barleycorns    make  1  Inch,  Used  by  shoemakers 

,  ^    -  «    w  TT     J         (Used    to  measure    the  height  or 

4  Inches  "    1  Hand,       \     .  .  ^-u     ^     ^A 

(     horses  at  the  shoulder. 

6  Feet  "    1  Fathom,      Used  to  measure  depths  at  sea. 

1.152|  Statute  Mi.  "    1  Geog.  mile,  Used  to  measure  distances  at  sea. 

3  Geographic    "    *'    1  League  =  3.458  st.  mi. 

60         '*  "    "I  1  D  i^^  latitude  on  a  meridian  or  of  longi- 

69.16  Statute     "     " )         ^^^^®  i     tude  on  the  equator. 

360  Degrees  **    the  Circumference  of  the  Earth. 

How  many  dimensions  has  extension  ?  Define  a  line.  Surface  or 
area.  A  solid  or  body.  Define  long  mensuro.  What  are  the  denom- 
inations?   The  value  of  each.    What  other  denominations  are  used? 


COMPOUND    KUMBERS. 


159 


1.  For  the  purpose  of  measuring  cloth  and  other  goods  sold  by  the  yard,  the  yard 
is  divided  into  halves,  fourths^  eigfUhs^  and  sixteenths.  The  old  table  of  cloth  meas- 
ure is  practically  obsolete. 

2.  The  geographic  mile  is  -^^  of  -ji^  or  ^^^qq  of  the  circumference  of  the  earth. 
It  is  a  small  fraction  more  than  1.15  statute  miles. 

3.  The  length  of  a  degree  of  latitude  varies,  being  68.72  miles  at  the  equator,  68.9 
to  69.05  miles  in  middle  latitudes,  and  69.30  to  69.34  miles  in  the  polar  regions.  The 
mean  or  average  length  is  as  stated  in  the  table.  A  degree  of  longitude  is  greatest 
at  the  equator,  where  it  is  69.16  miles,  and  it  gradually  decreases  toward  the  poles, 
where  it  is  0. 


Examples  for  Practice. 


1.  In  2  mi.  192  rd.  2  yd. 
how  many  inches  ? 


2.    In  164808  inches  how 
many  miles  ? 


OPERATION. 

operation. 

2  mi. 

192  rd.  2 

yd. 

12  )  164808  in. 

320 
832  rd. 

3 )  13734  ft. 
5J\4578yd. 
2  /   2 

416 

4162 

4578  yd. 
3 

11 )  9156 

32|0)83!2rd.  +  f  yd.=2yd. 
2  mi. +192  rd. 

13734  ft. 
12 

Ans.  2  mi.  192  rd.  2  yd. 

164808  in., 

Ans, 

3.  The  diameter  of  the  earth  being  7912  miles,  how  many 
inches  is  it  ?  Arts,  501304320  inches. 

4.  In  168474  feet  how  many  miles  ? 

5.  In  31  mi.  290  rd.  3  yd.  how  many  feet  ? 

6.  If  the  greatest  depth  of  the  Atlantic  telegraphic  cable 
from  l^ewfoundland  to  Ireland  be  2500  fathoms,  how  many 
miles  is  it  ?  Ans.  2  mi.  269  rd.  1|  ft. 

7.  If  this  cable  be  2200  miles  in  length,  and  cost  10  cents 
a  foot,  what  was  its  whole  cost?  Ans.  $1161600. 


160  REDUCTIONS'. 

8.  A  pond  of  water  measures  4  fathoms  3  feet  8  inches  in 
depth  ;  how  many  inches  deep  is  it  ?  Ans,  332. 

9.  How  many  times  will  the  driving  wheels  of  a  locomo- 
tive turn  round  in  going  from  Albany  to  Boston,  a  distance 
of  200  miles,  supposing  the  wheels  to  be  18  ft.  4  inches  in 
circumference  ?  Ans.  57600  times. 

10.  If  a  vessel  sail  120  leagues  in  a  day,  how  many  statute 
miles  does  she  sail?  Ans,  414.96  st.  mi. 

11.  How  many  inches  high  is  a  horse  that  measures  14J 
hands?  Ans.  5S. 

Surveyor's  Long  Measure. 
194,  A  Gunter's  Chain,  used  by  land  surveyors,  is  4. 
rods  or  66  feet  long,  and  consists  of  100  links. 

Table. 

7.92  Inches  (in.)  make  1  Link L 

25      Links  "     1  Rod rd. 

4      Rods,  or  66  feet,      «     1  Chain ch, 

80      Chains  "     1  Mile mi. 

Unit  Equivalents. 

1.  in. 

rd.  1  =     79.3 

ch.         1=     35=      198 

mi      1  =     4  =    100  =     793 

1  =  80  =  320  =  8000  =  63360 

Scale— ascending,  7.92,  25,  4,  80 ;  descending,  80,  4, 25,  7.93. 

Rods  are  seldom  used  in  chain  measure,  distances  being  taken  in  chains  and  hun         ; 
iredths.  ■ 

Examples  for  Practice.  1 

1.  In  3  mi.  61  ch.  73  1.  how  many  links? 

2.  Reduce  29173  1.  to  miles. 

3.  A  certain  field,  enclosed  by  a  board  fence,  is  17  ch. 
31 1.  long,  and  12  ch.  87 1.  wide  ;  how  many  feet  long  is  the 
fence  which  encloses  it  ?  Ans.  3983.76  ft. 

Repeat  the  ta])lp  of  surveyors*  long  measure.     Give  the  scale. 


COMPOUND    K  UMBERS. 


161 


12  in. 

= 

1 

fc 

■ 

1-! 

«-" 

II 

S 

f* 

- 

^ 

12  in.  =  1  ft. 


II.  Square  Measure. 

195,  A  Square  is  a  figure  bounded  by  four  equal  sides, 
and  having  four  right  angles. 

1  square  foot  is  a  figure  having 
four  sides  of  1  ft.  or  12  in.  each,  as 
shown  in  the  diagram.  Its  contents 
are  12  x  12  =  144  square  incheso 
Hence, 

The  contents  or  area  of  a  square, 
or  of  any  other  figure  having  a  uni- 
form length  and  a  uniform  hreadthy 
is  found  hy  7nultiplying  the  length  ly  the  breadth.  Thus,  a 
square  foot  is  12  in.  long  and  12  in.  wide,  and  the  contents 
are  12^  12  =  144  square  inches.  A  board  20  in.  long  and 
10  in.  wide  is  a  rectangle,  containing  20  x  10  =  200  square 
inches. 

196.  Square  Measure  is  used  in  computing  areas  or 
surfaces ;  as  of  land,  boards,  painting,  plastering,  pa^dng,  etc. 

Table. 

144  Square  Inches  (sq.  in.)  make  1  Square  Foot sq.  ft. 

9  Square  Feet  "     1  Square  Yard sq.  yd. 

30|^  Square  Yards  '*     1  Square  Rod sq.  rd. 

160  Square  Rods  "    1  Acre A. 

640  Acres  **    1  Square  Mile sq.  mi. 


Unit  Equivalents. 


A. 
sq.mi.     1  = 


sq.  rd. 
1  = 

160  = 


eq.  yd. 

1  = 

30i  = 

4840  = 


sq.  ft. 
1  = 
9  = 

2721  = 
43560  = 


sq.  m, 

144 

1296 

39204 

6272640 


1  =  640  =  102400  =  3097600  =  27878400  =  4014489600 

Scale— ascending,  144,  9,  30|,  160,  640 ;  descending.  640,  160,  30^, 
9,144. 

Define  a  square.  How  is  tlie  area  of  a  square  or  any  rectangular 
figure  found  ?  For  what  is  square  measure  used  ?  Repeat  the  table. 
Give  the  scale. 


162  REDUCTION. 

Artificers  estimate  fheir  work  as  follows  : 

By  the  square  foot ;  glazing  and  stone-cutting. 

By  the  square  yard  ;  painting,  plastering,  paving,  ceiling,  and  paper- 
hanging. 

By  the  square  of  100  feet ;  flooring,  partitioning,  roofing,  slating, 
and  tiling. 

Brick-work  is  generally  estimated  by  the  1000  bricks ;  sometimes  m 
yubic  feet. 

1.  In  estimating  the  painting  of  moldiogs,  cornices,  etc.,  the  measuring-line  Is 
earried  into  all  the  moldings  and  cornices. 

2.  A  brick  wall  which  is  a  brick  and  a  half  thick,  is  said  to  be  of  the  standard 
thickness.    Five  courses  in  the  height  are  called  a  foot. 

Examples  for  Practice. 

1.  In  10  A.  65  sq.  rd.  16  sq.  yd.  4  sq.  ft.  136  sq.  in.  how 
many  square  inches  ?  • 

OPERATION. 

10  A.  65  sq.  rd.  16  sq.  yd.  4  sq.  ft.  136  sq.  in. 
160 
1665  sq.rd. 

30} 

416} 
49966 
50382}  sq.  yd. 

9_ 

453444}  sq.  ft. 
144 


36  =  }  sq.  ft. 
1813912  with  136  sq.  in. 
1813776 
453444 
65296108  sq.  in.,  Ans, 

%  In  65296108  sq.  in.  how  many  acres  ? 


How  do  artisans  estimate  work  ? 


COMPOUND     Is^UMBERS.  163 

OPERATION. 
144 )  65296108  sq.  in. 

9 )  453445  sq.  ft.  +  28  sq.  in. 
son  50382  sq.  yd.  +  7  sq.  ft. 

4  J         4 


121 )  201528  fourths  sq.  yd. 

16|0U66i5  sq.  rd.  +  ^=15|  sq.  yd. 

10  A.  +  65.  sq.  rd. 
Ans.  10  A.  65  sq.  rd.  15f  sq.  yd.  7  sq.  ft.    28  sq.  in. 

ilO  A.  65  sq.  rd.  15    sq.  yd.  7  sq.  ft.    28  sq.  in. 
6  sq.  ft.  108  sq.  in. 

Or       10  A.  65  sq.  rd.  16    sq.  yd.  4  sq.  ft.  136  sq.  in. 

Analysis.  Dividing  by  the  numbers  in  the  ascending  scale,  and 
arranging  the  remainders  according  to  their  order  in  a  line  below,  we 
find  the'fequare  yards  a  mixed  number,  15|.  But  f  of  a  sq.  yd.  =  |  of 
9  sq.  ft.  =  6|  sq.  ft. ;  and  |  of  a  sq.  ft.  =  f  of  144  sq,  in.  =  108  sq.  in. 
Therefore  |  sq.  yd.  =  6  sq.  ft.  108  sq.  in. ;  and  adding  108  sq.  in.  to 
28  sq.  in.  we  have  136  sq.  in. ,  and  6  sq.  ft.  to  7  sq.  ft.  we  have  13  sq. 
ft.  =  1  sq.  yd.  4  sq.  ft.,  and  writing  the  4  sq.  ft.  in  the  result,  and 
adding  1  sq.  yd.  to  15  sq.  yd.  we  have  for  the  reduced  result,  10  A. 
65  sq.  rd.  16  sq.  yd.  4  sq.  ft.  136  sq.  in. 

3.  Reduce  87  A.  118  sq.  rd.  7  sq.  yd.  1  sq.  ft.  100  sq.  in. 
to  square  inches.  Ans.  550355068  sq.  in. 

5.  Reduce  550355068  square  inches  to  acres. 

5.  A  field  100  rods  long  and  30  rods  wide  contains  how 
many  acres  ?  Ans,  18  A.  120  sq.  rd. 

6.  How  many  rods  of  fence  will  enclose  a  farm  a  mile 
square  ?  Ans.  1280  rods. 

7.  How  much  additional  fence  wiU  divide  it  into  four 
equal  square  fields  ?  Ans.  640  rods. 

8.  How  many  acres  of  land  in  Boston,  at  $1  a  square 
foot,  ^vill  $100000  purchase  ? 

Ans.  2  A.  47  sq.  rd.  9  sq.  yd.  3^  sq.  ft. 

9.  How  many  yards  of  carpeting,  1  yd.  wide,  will  be  re- 
quired to  carpet  a  room  18^  ft.  long  and  16  ft.  wide  ? 

Ans.  32|^  yards. 


164  KEDUCTIOI^r. 

10.  What  would  be  the  cost  of  plastering  a  room  18  ft. 
long,  16 J  ft.  wide,  and  9  ft.  high,  at  22  cents  a  sq.  yd.  ? 

Ans.  $22.44. 

11.  What  will  be  the  expense  of  slating  a  roof  40  feet 
long  and  each  of  the  two  sides  20  feet  wide,  at  $10  per 
square?  Ans.  $160. 

Surveyors'  Square  Measure. 

197.  This  measure  is  used  by  surveyors  in  computing  the 
Area  or  contents  of  land. 

Table. 
635  Square  Links  (sq.  1.)  make  1  Pole P. 

16  Poles  "      1  Square  Chain sq.  ch. 

10  Square  Chains  *'      1  Acre A. 

640  Acres  "     1  Square  Mile sq,  mi. 

86  Square  Miles  (6  miles  square)  "      1  Township Tp. 

Unit  Equivalents. 

P.  sq.  1. 

sq.  ch.  1  =  625 

A.  1  =  16  =  10000 

sq.  mi.         1  =         10  =         160  =         100000 

Tp.       1  =      640  =      6400  =    103400  =      64000000 

1  =  36  =  23040  =  330400  =  3686400  =  3304000000 

Scale— ascending,  635,  16,  10,  640,  36  ;  descending,  36,  640,  10, 16, 

635. 

1.  A  pqnare  mile  of  land  is  also  called  a  section. 

2.  Engineers  commonly  use  a  chain,  or  measuring  tape  100  feet  long,  each  foot 
divided  into  tenths. 

3.  The  contents  of  land  are  commonly  estimated  In  square  miles,  acres,  and  hun- 
dredths ;  the  denomination,  rood,  is  no  longer  used. 

Examples  for  Practice. 

1 .  How  many  poles  in  a  township  of  land  ? 

2.  Eeduce  3686400  P.  to  sq.  mi. 

3.  In  94  A.  7  sq.  ch.  12  P.  118  sq.  1.  how  many  square 
links  ? 

4.  What  will  be  the  cost  of  a  farm  containing  4550000 
square  links,  at  $50  per  acre  ?  Ans.  12275. 

Repeat  the  table  of  surveyors'  square  measure.     Give  the  scale. 


compou:n"I)    numbers. 


165 


III.    Cubic  Measure. 

198.  A  Cube  is  a  solid,  or  body, 

ba\ing  six  equal  square  sides,   or 

faces.    If  each  side  of  a  cube  be  1 

yard,  or  3  feet,  1  foot  in  thickness 

of  this  cube  will  contain  3x3x1 

=9  cubic  feet,  and  the  whole  cube 

will  contain  3  x3  x  3  =  27  cubic 

feet. 

A  solid,  or  body,  may  have  the  three  dimensions  all  alike 

or  all  different.     A  body  4  ft.  long,  3  ft.  wide,  and  2  ft.  thick 

contains  4  x  3  x  2  =  24  cubic  or  solid  feet.    Hence, 

The  cubic  or  solid  contents  of  a  hody  are  found  hy  muUi- 
plying  the  lengtlh  breadth,  and  thickness  together, 

199,  Cubic  Measure,  also  called  Solid  Measure,  is  used 
in  estimatmg  the  contents  of  solids,  or  bodies ;  as  timber, 
wood,  stone,  etc. 

Table. 
1728  Cubic  Inches  (cu.  in.)  make  1  Cubic  Foot en.  ft. 


27  Cubic  Feet 
16  Cubic  Feet 
8  Cord  Feet,  or 
128  Cubic  Feet, 

24^  Cubic  Feet 


1  Cubic  Yard ... cu.  yd. 

1  Cord  Foot cd.  ft. 

1  Cord  of  Wood .Cd. 


j  Perch  of  Stone ) 
\  or  Masonry,       J 


,.Pch. 


Scale — ascending,  1728, 27.  The  other  numbers  are  not  in  a  regular 
scale,  but  are  merely  so  many  times  1  foot.  The  unit  equivalents,  be- 
ing fractional,  are  consequently  omitted. 

1.  A  cubic  yard  of  earth  la  called  a  load, 

2.  Railroad  and  transportation  companies  estimate  light  freight  by  the  ^pace  it 
occupies  in  cubic  feet,  and  heavy  freight  by  weight. 

3.  A  pile  of  wood  8  feet  long,  4  feet  wide,  and  4  feet  high,  contains  1  cord ;  and  a 
cord  foot  is  1  foot  in  length  of  such  a  pile. 

4.  A  perch  of  stone  or  of  masonry  is  16  j  feet  long,  1^  feet  wide,  and  1  foot  high. 

Define  a  cube.  How  are  the  contents  of  a  cube  or  rectangular  solid 
found  ?  For  what  is  cubic  measure  used  ?  Repeat  the  table.  Give 
the  scale.  How  is  railroad  freight  estimated  ?  What  is  understood 
by  a  cord  foot  ?    By  a  perch  of  stone  or  masonrj'  ? 


166  REDUCTIOlf. 

6.  Joiners,  bricklayers,  and  masons  make  no  allowance  for  windows,  doors,  etc. 

Bricklayers  and  masons,  in  estimating  their  work  by  cubic  measure,  make  no  allow- 
ance for  the  corners  of  the  walls  of  houses,  cellars,  etc.,  but  estimate  their  worK  by 
the  girt,  that  is,  the  entire  length  of  the  wall  on  the  outside. 

6.  Engineers,  in  making  estimates  for  excavations  and  embankments,  take  the 
dimensions  with  a  line  or  measure  divided  into  feet  and  decimals  of  a  foot.  Tho 
estimates  are  made  in  feet  and  decimals,  and  the  results  are  reduced  to  cubic  yards. 

Examples  for  Practice. 

1.  In  125  cu.  ft.  840  cu.  in.  how  many  cu.  in.  ?  Ans,  216840. 

2.  Eeduce  5224  cubic  feet  to  cords.  A7is.  40|f. 

3.  In  a  solid,  3  ft.  2  in.  long,  2  ft.  2  in.  wide,  and  1  ft. 
8  in.  thick,  how  many  cubic  inches?  Ans.  19760. 

4.  How  many  small  cubes,  1  inch  on  each  edge,  can  be 
sawed  from  a  cube  6  feet  on  each  edge,  allowing  no  waste 
for  sawing?  Ans,  37324:8. 

6.  In  a  pile  of  wood  60  feet  long,  20  feet  wide,  and  15  feet 
high,  how  many  cords  ?  Ans.  140|. 

6.  How  many  cubic  feet  in  a  load  of  wood  10  feet  long,  3;^ 
feet  wide,  and  3^  feet  high  ?  Ans.  113f  cu.  ft. 

7.  If  a  load  of  wood  be  12  feet  long  and  3  feet  wide,  how 
high  must  it  be  to  make  a  cord  ?  Ans.  3^  ft.  high. 

8.  The  gray  limestone  of  Central  New  York  weighs  175 
pounds  a  cubic  foot.     What  is  the  weight  of  one  solid  yard  ? 

Ans.  2  T.  7  cwt.  25  lb. 

9.  A  cellar  wall,  32  ft.  by  24  ft,  is  6  ft.  high  and  1  ^  ft.  thick. 
What  did  it  cost  at  $1.25  a  perch?  A}is.  $50,909  +  . 

10.  What  did  it  cost  to  dig  the  same  cellar,  at  15  cents  a 
cubic  yard?  Ans,  $25.60. 

11.  My  sleeping  room  is  10  ft.  long,  9  ft.  wide,  and  8  ft. 
high.  If  I  breathe  10  cu.  ft.  of  air  in  one  minute,  in  how 
long  a  time  will  I  breathe  as  much  air  as  the  room  contains  ? 

Ans.  72  min. 

12.  In  a  school-room  30  ft.  long,  20  ft.  wide,  and  10  ft.  high, 
with  50  persons  breathing  each  10  cu.  ft.  of  air  in  one  minute, 
in  how  long  a  time  will  they  breathe  as  much  as  the  room 
contains?  Ans.  12  min. 

How  are  excavations  and  embankments  measured  ? 


COMPOUND    NUMBERS.  167 

MEASURES    OF    CAPACITY. 

I.    Liquid  Measuee. 

200.  Liquid  Measure,  also  called  Wine  Measure,  is 
used  in  measuring  liquids  ;  as  liquors,  molasses,  water,  etc. 

Table. 

4   Gills  (gl.)  make  1  Pint pt. 

3  Pints  "      1  Quart qt. 

4  Quarts  "      1  GaUon gaL 

8H  Gallons  "      1  Barrel bbl. 

3    Barrels,  or  63  gal.     "      1  Hogshead hlid. 

Unit  Equivalents. 

pt.         gL 

qt.         1=4 

gal.  1=2=       8 

bbl.      1    =      4=      8=     82 

hhd.  1  =  31i  =  126  =  252  =  1008 

1  =  2  =  63  =  252  =  504  =  2016 

Scale— ascending,  4,  2,  4,  31|,  3 ;  descending,  2,  31|^,  4,  2,  4. 

The  following  denominations  are  also  in  use  : 

36  Gallons  make  1  Barrel       of  beer. 

54       *'        oi-t^  Barrels  "      1  Hogshead  "     " 

42       "  *•      1  Tierce. 

2  Hogsheads,  or  130  gallons,      "      1  Pipe  or  Butt 

2  Pipes,  or  4  Hogsheads,  "      1  Tun. 

1.  The  denominations,  barrel  and  hogshead,  are  used  in  estimating  the  capacity 
of  cisterns,  reservoirs,  vats,  etc. 

2.  The  tierce,  hogshead,  pipe,  butt,  and  tun  are  the  names  of  casks,  and  do  not 
express  any  fixed  or  definite  measures.  They  are  usually  gauged,  and  have  their 
capacities  in  gallons  marked  on  them. 

3.  Ale  or  beer  measure,  formerly  used  in  measuring  beer,  ale,  and  milk,  is  almost 
entirely  discarded. 

What  is  liquid  measure  ?  Repeat  the  table.  Give  the  scale.  What 
other  denominations  are  sometimes  used  ?  How  are  the  capacities  of 
cisterns,  reservoirs,  etc.,  reckoned  ?    Of  large  casks  ? 


168 


REDUCTION. 


Examples  for  Practice. 


1.  In  2  hhd.  1  bar.  30  gal  2 
qt.  1  pt.  3  gi.  how  many  gills  ? 

OPERATION. 

2  hhd.  1  bar.  30  gal.  2  qt. 
_2  [1  pt.  3  gi. 

5bbl. 
31i 

185 


187i  gal- 
4 

752  qt. 

2 


1505  pt. 
4 


2.  In  6023  gi.  how  many 
hhds.  ? 

OPERATION. 

4 )  6023  gi. 
2)  1505pt.+3gi. 
4)  752qt.  +  lpt. 


31i 

2 


188  gal. 

2 


63  )  376 


[gal. 
=  30J 


2)_5bbl.-f^gal. 

2  hhd. +  1  bar. 
Ans,  2  hhd.  1  bar.  30J  gal. 
1  pt.  3  gi. 

But  -J  gal.  =  2  qt.,  making 
the  Ans.  2  hhd.  1  bar.  30  gal. 
6023  gi.,  Ans.  2  qt.  1  pt.  3  gi. 

3.  Eeduce  3  hogsheads  to  gills. 
"  4.  Reduce  6048  gills  to  hogsheads. 
6.  In  13  hhd.  15  gal.  1  qt.  how  many  pints  ? 

6.  In  6674  pints  how  many  hogsheads  ? 

7.  What  wiU  be  the  cost  of  a  hogshead  of  wine,  at  6  cents 
a  gill?  Ans.  $120.96. 

8.  A  grocer  bought  10  barrels  of  cider,  at  $2  a  barrel ; 
after  converting  it  into  vinegar,  he  retailed  it  all  at  5  cents 
a  quart ;  what  was  his  whole  gain  ?  Ans,  $43. 

9.  At  6  cents  a  pint,  how  much  molasses  can  be  bought 
for  $3.84?  A71S,  8  gal. 

10.  How  many  demijohns,  that  will  contain  2  gal.  2  qt.  1  pt 
each,  can  be  fiUed  from  a  hogshead  of  wine  ?       Ans.  24. 

II.    Dry  Measure. 
201.   Dry  Measure  is  used  in  measuring  articles  not 
liquid,  as  grain,  fruit,  salt,  roots,  allies,  etc. 

What  is  dry  measiL'e  ? 


COMPOUND    NUMBERS.  169 

Table. 

3  Pints  (pt.)  make  1  Quart qt. 

8  Quarts  "     1  Peck pk. 

4  Pecks  "     1  Bushel bu. 

Unit  Equivalents. 
qt,      pt. 
pk.       1=3 
bu.      1  =    8  =  16 
1  =  4  =  32  =  64 
Scale— ascending,  3,  8,  4 ;  descending,  4,  8,  2. 
In  England,  8  bu.  of  70  lbs.  each  are  called  a  quarter,  used  in  measaring  grain. 
The  weight  of  the  English  quarter  is  ^  of  a  long  ton. 

Examples  for  Practice. 

1.  In  49  bu.  3  pk.  7  qt.  1  pt.  how  many  pints  ? 

2.  In  3199  pt.  how  many  bushels  ? 

3.  Reduce  1  bu.  1  pk.  1  qt.  1  pt.  to  pints. 

4.  Reduce  83  pints  to  bushels. 

5.  An  innkeeper  bought  a  load  of  50  bushels  of  oats  at 
65  cents  a  bushel,  and  retailed  them  at  25  cents  a  peck ; 
how  much  did  he  make  on  the  load?  Ans,  $17.50. 

Stan^dard  of  Exten^siok. 

202.  TJie  U.  8.  standard  unit  of  measures  of  extension, 
whether  linear,  superficial,  or  solid,  is  the  yard  of  3  feet,  or 
36  inches,  and  is  the  same  as  the  imperial  standard  yard  of 
Great  Britain.  It  is  determined  as  follows  :  The  rod  of  a 
pendulum  vibrating  seconds  of  mean  time,  in  the  latitude 
Oi  London,  in  a  vacuum,  at  the  level  of  the  sea,  is  divided 
into  391393  equal  parts,  and  360000  of  these  parts  are  36 
inches,  or  1  standard  yard.  Hence,  such  a  pendulum  rod 
is  39.1393  inches  long,  and  the  standard  yard  is  |§;§^§  of 
the  length  of  the  pendulum  rod. 

203.  Hie  U.  S.  standard  unit  of  liquid  measure  is  the 
old  English  wine  gallon,  of  231  cubic  inches,  which  is  equal 
to  8.33888  pounds  avoirdupois  of  distilled  water  at  its  maxi- 
mum density,  that  is,  at  the  temperature  of  39.83°  Fahren- 
heit, the  barometer  at  30  inches. 

Repeat  the  table.  What  is  a  quarter?  Wliat  is  the  TJ.  S.  standard 
unit  of  measurement  of  extension  ?  How  is  it  determined  ?  What  is 
the  U.  S.  standard  unit  of  liquid  measure  ? 

n-  P.  8 


170  REDUCTION^. 

304.  The  U.  8.  standard  unit  of  dry  measure  is  the  Brit- 
fsli  Winchester  bushel,  which  is  18^  inches  in  diameter  and 
8  inches  deep,  and  contains  2150.42  cubic  inches,  equal  to 
77.6274  pounds  avoirdupois  of  distilled  water,  at  its  maximum 
density.  A  gallon,  dry  measure,  contains  268.8  cubic  inches. 

1.  The  wine  and  dry  measures  of  the  same  denomination  are  of  different  capaci- 
ties.   The  exact  and  the  relative  size  of  each  may  he  readily  seen  hy  the  following 

205.    COMPARATIVB  TABLE  OP  MEASURES  OP  CAPACITY. 

Cu.  in.  in        On.  in.  in       Cu.  in.  in        Cu.  in.  in 
one  gallon.       one  quart.       one  pint  one  gill. 

Wine  measure,  231  57|  28|  1^ 

Dry  measure,  Q  pk.,)  268^  67^  33f  8f 

2.  The  beer  gallon  of  282  inches  is  retained  in  use  only  by  custom.  A  bushel  is 
commonly  estimated  at  2150.4  cubic  iaches. 

Examples  for  Practice. 

1.  A  fruit  dealer  bought  a  bushel  of  strawberries,  dry 
measure,  and  sold  them  by  wine  measure  ;  how  many  quarts 
did  he  gain  ?  Ans,  5^  quarts. 

2.  A  grocer  bought  40  quarts  of  milk  by  beer  measure, 
and  sold  it  by  wine  measure ;  how  many  quarts  did  he  gain  ? 

Ans.  844"  quarts. 

3.  A  bushel,  or  32  quarts,  dry  measure,  contains  how 
many  more  cubic  inches  than  32  quarts  wine  measure  ? 

A71S.  302|  cu.  in. 
Time. 

306.  Time  is  used  in  measuring  periods  of  duration,  as 

years,  days,  minutes. 

Table. 

60  Seconds  (sec.)      make  1  Minute min. 

GO  Minutes  "     1  Hour h. 

24Hours  **     1  Day da. 

7  Days  "     1  Week wk. 

365  Days  **     1  Common  Year yr. 

866  Days  '*     1  Leap  Year yr. 

13  Calendar  Months    **     1  Year yr. 

100  Years  **     1  Century C. 

What  is  the  U.  S.  standard  unit  of  dry  measure?  How  is  it  ob- 
toined?  What  is  the  relative  size  of  the  wine  and  the  dry  gallon  t 
What  is  the  size  of  a  beer  gallon  ?    What  is  time  ?    Repeat  the  table. 


COMPOUNDS  UMBERS.  171 


Unit  Equivat<f,nt8. 

min. 

sec 

h.              1  = 

60 

da.           1  =         60  = 

8600 

wk.             1  =      24  =      1440  = 

86400 

1  =          7  =    168  =    10080  = 

604800 

mo.       (  865  =  8760  =  525600  = 
12  =  "j  366  =  8784  =  527040  = 

31536000 

31622400 

Season. 

Names, 

Abbreviations. 

Winter, 

(   Istn 
\  2d 

lonth 

,  January, 

Jan. 

February, 

Feb. 

(    3d 

March, 

Mar. 

Spring, 

■j    4th 

April, 

Apr. 

(    5th 

May, 

May 

(    6th 

June, 

Jun. 

Summer, 

\    7th 

July, 

July 

(    8th 

August, 

Aug. 

9th 

September, 

Sept. 

Autumn, 

■  10th 

(nth 

October, 

Oct. 

November, 

Nov. 

Winter, 

12th 

December, 

Dec. 

Scale— ascending,  60,  60,  24,  7  ;  descending,  7,  24,  60,  60. 

The  calendar  year  is  divided  as  follows  ; 

No.  of  days. 
31 

28  or  29 
31 
30 
31 
30 
31 
31 
30 
31 
30 
31 

365  or  366 

1.  The  exact  length  of  a  solar  year  is  365  da.  5  h.  48  min.  46  sec. ;  but  for  conve- 
nience it  is  reclioned  11  min.  14  sec.  more  than  this,  or  365  da.  6h.  =  365j  da.  This  \ 
day  in  4  years  makes  one  day,  which,  every  fourth,  bissextile,  or  leap  year,  is  added 
to  the  shortest  month,  giving  it  29  days.  The  leap  years  are  exactly  divisible  by  4, 
as  1856, 1860, 1864.  The  number  of  days  in  each  calendar  month  may  be  easily  re- 
membered by  committing  the  following  lines : 

''  Thirty  days  hath  September, 
April,  JunCj  and  November  ; 
All  the  vest  have  thirty-one. 
Save  February,  which  alone 
Hath  twenty-eight ;  and  one  day  more 
We  add  to  it  one  year  in  four." 

2,  In  most  business  transactions  30  days  are  called  1  month. 

Examples  fok  Pkactice. 

1.  Reduce  365  da^  5  h.  48  min.  46  sec.  to  seconds. 

2.  Reduce  31556926  seconds  to  days. 

Give  the  scale.  What  is  the  length  of  each  of  the  calendar  months  ? 
What  is  the  exact  length  of  a  solar  year?  Explain  the  use  of  bissextile 
or  leap  year.    What  is  the  length  of  a  month  in  business  transactions  t 


172  REDUCTION. 

3.  In  5  wk.  1  da.  1  h.  1  min.  1  sec.  how  many  seconds  ? 

4.  In  3114061  seconds  how  many  weeks  ? 

5.  How  many  times  does  a  clock  pendulum,  3  ft.  3  in. 
long,  beating  seconds,  vibrate  in  one  day?      Ans.  86400. 

6.  If  a  man  take  1  step  a  yard  long  in  a  second,  in  how 
long  a  time  will  he  walk  10  miles?  Ans.  4  h.  53  min.  20  sec. 

7.  In  a  lunar  month  of  29  da.  12  h.  44  min,  3  sec.  how 
many  seconds?  Ans.  2551443. 

8.  How  much  time  will  a  person  gain  in  40  years,  by  rising 
45  minutes  earlier  every  day?  Ans.  456  da.  13  h.  30  min. 

Circular  Measure. 

307.  Circular  Measure,  or  Circular  Motion,  is  used 
principally  in  surveying,  navigation,  astronomy,  and  geogra- 
phy, for  reckoning  latitude  and  longitude,  determining  loca- 
tions of  places  and  vessels,  and  computing  difference  of  time. 
Every  circle,  great  or  small,  is  divisible  into  the  same  num- 
ber of  equal  parts,  as  quarters,  called  quadrants,  twelfths, 
called  signs,  360ths,  called  degrees,  etc.  Consequently  the 
parts  of  different  circles,  although  having  the  same  names, 
are  of  different  lengths. 

Table. 

60  Seconds  (' )         make  1  Minute '. 

60  Minutes  "     1  Degree °. 

30  Degrees  "     1  Sign S. 

12  Signs,  or  360°,        "    1  Circle C. 

Unit  Equivalents. 

1=  60 

S.         1  =       60  =       8600 

C.        1  =    30  =    1800  =    108000 

1  =  12  =  380  =  21600  =  1296000 

Scale— ascending,  60,  60,  30,  13  ;  descending,  12,  80,  60,  60. 

1.  Minutes  of  the  earth's  circumference  are  called  geographir  or  nautical  miles. 

2.  The  denomination,  signs,  is  confined  exclusively  to  Astronomy. 

Define  circular  measure.      How  are  circles  divided  ?    Repeat  the 
table.     Give  the  scale.    What  is  a  geographic  mile  ?    What  is  a  sign  ? 


COMPOUN^D    NUMBERS. 


173 


3.  Degrees  are  not  strictly  divisions  of  a  circle,  but  of  the  space  about  a  point  in 
any  plane. 

4.  90°  make  a  quadrant,  or  right  angle,  and  60°  a  sextant,  or  |  of  a  circle. 

Examples  for  Practice. 

1.  Eediice  10  S.  10°  10'  10"  to  seconds. 

2.  Eeduce  1116610"  to  signs. 

3.  How  many  degrees  in  11400  geographic  vt  nautical 
miles?  ^ws.  190^ 

4.  If  1  degree  of  the  earth's  circumference  is  69^  statute 
miles,  how  many  statue  miles  in  11400  geographic  miles,  or 
190  degrees  ?  Ans.  13148. 

5.  How  many  minutes,  or  nautical  miles,  in  the  circum' 
ference  of  the  earth  ?  Ans.  21600'  or  mi. 

6.  A  ship  during  4  days'  storm  at  sea  changed  her  longi- 
tude 397  geographic  miles;  how  many  degrees  and  minutes 


did  she  change  ? 

Ans.  6°  37'. 

308.    Iif  CouinTting. 

13  Units  or  Things make 

! 1  Dozen. 

12  Dozen 

1  Gross. 

12  Gross 

1  Great  Gross. 

20  Units 

1  Score. 

209.     Paper. 

24  Sheets.. make 

....1  Quire. 

20  Quires                   " 

1  Ream. 

2  Reams 

1  Bundle. 

5  Bundles                 ** 

IBale. 

210.    Books. 

The  terms  folio,  quarto,  octavo, 

duodecimo,  etc.,  indicate 

the  number  of  leaves  into  which  a  i 

sheet  of  paper  is  folded. 

A  sheet  folded  in    2  leaves  is  called  a  Folio. 

A  sheet  folded  in    4  leaves       ** 

a  Quarto,  or  4to. 

A  sheet  folded  in    8  leaves       " 

an  Octavo,  or  8vo. 

A  sheet  folded  in  12  leaves       ** 

a  12mo. 

A  sheet  folded  in  16  leaves       *' 

a  16mo. 

A  sheet  folded  in  18  leaves       " 

an  18mo. 

A  sheet  folded  in  24  leaves 

a24mo. 

A  sheet  folded  in  32  leaves       " 

a  82mo. 

What  is  a  degree?     Repeat  the  table  for  counting.     For  reckoning 
paper.     For  indicating  the  size  of  books. 


174  REDUCTIOiJ^. 

Examples  for  Practice. 

1.  If  in  Birmingham,  England,  150  million  Gillott  pens  are 
manufactured  annually,  how  many  great  gross  will  they 
make  ?  Ans.  86805  great  gross  6  gross  8  dozen. 

2.  In  100000  sheets  of  paper,  how  many  bales  ? 

A71S.  20  bales  4  bundles  6  quires  16  sheets. 

3.  What  is  the  age  of  a  man  4  score  and  10  years  old  ? 

4.  How  many  printed  pages,  2  pages  to  each  leaf,  will 
there  be  in  an  octavo  book,  having  8  fully  printed  sheets  ? 

Ans.  128  sheets. 

5.  How  large  a  book  will  ten  32mo  sheets  make,  if  every 
page  be  printed  ?  Ans.  640  pages. 

Promiscuous  Examples  in  Eeduction". 

1.  How  many  suits  of  clothes,  each  containing  6  yd.  Bj  qr., 
can  be  cut  from  333  yards  of  cloth  ?  Ans.  48. 

2.  A  man  bought  a  gold  chain,  weighing  1  oz.  15  pwt.,  at 
seven  dimes  a  pennyweight ;  what  did  it  cost?    Ans.  $24.50. 

3.  A  physician,  having  2  ft)  3  ^  5  3  1 3  10  gr.  of  medicine, 
dealt  it  out  in  prescriptions  averaging  15  grains  each  ;  how 
many  prescriptions  did  it  make  ?  A7is.  886. 

4.  A  man  bought  IT.  11  cwt.  12  lbs.  of  hay,  at  IJ  cents 
a  pound  ;  what  did  it  cost  ?  Ans.  $38.90. 

5.  What  will  be  the  cost  of  a  load  of  oats  weighing  1456 
pounds,  at  37J-  cents  per  bushel?  Atis,  $17.0625. 

6.  If  one  bushel  of  wheat  will  make  45  pounds  of  flour,  how 
many  barrels  will  1000  bushels  make  ?    Ans.  229  bbl.  116  lb. 

7.  A  load  of  wheat  weighing  2430  pounds  is  worth  how 
much,  at  $1.20  a  bushel?  Ans.  $48.60. 

8.  Paid  $12.50  for  a  barrel  of  beef  ;  how  much  was  that 
per  pound?  Ans.  6 J  cents. 

9.  If  a  silver  dollar  measure  one  inch  in  diameter,  how 
many  dollars,  laid  side  by  side  on  the  equator,  would  reach 
round  the  earth  ?  A?is.  1577511936. 

10.  In  10  mi.  7  ch.  4  rd.  20  1.,  how  many  links  ? 

Ans.  808:30  links. 


DEIfOMINATE     FRACTIONS.  175 

11.  What  is  the  value  of  a  city  lot,  25  feet  wide  and  100 
feet  long,  if  every  square  inch  is  worth  one  cent  ?  Ans.  $3 GOO. 

12.  How  many  cords  of  wood  can  be  piled  in  a  shed  50  ft. 
long,  25  ft.  wide,  and  10  ft.  high  ?  Ans.  97  Cd.  5  cd.  ft.  4  cu.  ft. 

13.  A  cistern  10  feet  square  and  10  feet  deep,  will  hold 
how  many  hogsheads  of  water  ?       Ans.  118  hhd.  464-^  gal. 

14.  A  bin  8  feet  long,  5  feet  wide,  and  4^  feet  high,  will 
hold  how  many  bushels  of  grain  ?  Ans.  144^^  bu. 

15.  How  many  seconds  less  in  every  Autumn  than  in 
either  Spring  or  Summer  ?  Ans.  86400  sec. 

16.  If  a  person  could  travel  at  the  rate  of  a  second  of  dis- 
tance in  a  second  of  time,  how  much  time  would  he  require 
to  travel  round  the  earth  ?  Ans.  15  days. 

17.  How  many  yards  of  carpeting,  1  yd.  wide,  will  be  re- 
quired to  carpet  a  room  20  ft.  long  and  18  ft.  wide  ? 

Ans.  40. 

18.  A  printer  calls  for  4  reams  10  quires  and  10  sheets  of 
paper  to  print  a  book  ;  how  many  sheets  does  he  call  for  ? 

Ans.  2170. 

19.  How  many  times  will  a  wheel,  16  ft.  6  in.  in  circum- 
ference, turn  round  in  running  42  miles  ?        Ans.  13440. 

20.  How  many  days,  working  10  hours  a  day,  will  it  re- 
quire for  a  person  to  count  110000,  at  the  rate  of  one  cent 
each  second  ?  Ans.  27  da.  7h.  46  min.  40  sec. 

21.  A  town,  6  miles  long  and  4^  miles  wide,  is  equal  to 
how  many  farms  of  80  acres  each?  Ans.  216. 

22.  At  $21.75  per  rod,  what  will  be  the  cost  of  grading 
10  mi.  176  rds.  of  road  ?  Ans.  $73428. 

REDUCTION  OF  DEISTOMINATE  FRACTIONS. 

Case  I. 
211.   To   reduce  a  denominate  fraction  from  a 
greater  to  a  less  unit. 

1.  Reduce  -^  of  a  bushel  to  the  fraction  of  a  piiit. 
Case  I  is  what  1 


176  REDUCTION. 

OPERATION.  Analysis.    To  reduce  bushels 

bn.                                  pt.  to  pints,  we  must  multiply  by  4, 

^Xfxfxf  =  |,  Ans.  8,  and  2,  tbe  numbers   in  the 

Or,  scale.    And  since  the  given  num- 


5 


1 


4  =  ^  pt.,  Ans. 


ber  is  a  fraction  of  a  bushel,  we 
indicate  the  process  as  in  multi- 
plication of  fractions,  and  after 
canceling,  obtain  |. 


iluLE.  Multiply  the  fraction  of  the  higher  denomination 
hy  the  numbers  in  the  scale  successively,  between  the  given 
and  the  required  denominations. 

Cancellation  may  be  applied  wherever  practicable. 

Examples  foe  Practice. 

2.  Reduce  xoW  of  a  £  to  the  fraction  of  a  penny. 

Ans.  -^  d. 

3.  Reduce  xihr  of  a  week  to  the  fraction  of  a  minute. 

Ans.  -^  min. 

4.  What  part  of  a  gill  is  ^^5^3^  of  a  hogshead  ?    Ans.  i  gi. 

5.  What  fraction  of  a  grain  is  -^q  of  an  ounce?  Ans.  ^  gr. 

6.  Reduce  ^Q^)i^ft^)  of  a  mile  to  the  fraction  of  an  inch  ? 

Ans.  -gS^in. 

7.  Reduce  |  of  ^  of  2  pounds  to  the  fraction  of  an  ounce 
Troy.  Ans.  f  oz. 

8.  Reduce  -^  of  a  hogshead  to  the  fraction  of  a  pint. 

Ans.  fl  pt. 

9.  Reduce  y^^  of  an  acre  to  the  fraction  of  a  rod. 

Ans.  i  rd. 

Case  II. 

212.  To  reduce  a  denominate  fraction  from  a  less 
to  a  greater  unit. 

1.  Reduce  f  of  a  pint  to  the  fraction  of  a  bushel. 

Give  explanation.    Rule.    Case  IT  ia  what  ? 


DENOMINATE    FRACTIONS.  177 

OPERATION.  Analysis.     To  reduce  pints 

^1111  to  bushels,   we  must  divide 

^   ^'2~^~g~^'4"^^  30^  ^^'      by  2,  8,  and  4,  the  numbers 

of  the  scale.  And  since  the 
given  number  of  pints  is  a 
fraction,  we  indicate  the  pro- 
cess, as  in  division  of  fractions, 
and  canceling,  obtain  -^-q. 


Or,     5 
2 

8 

4 


80     1  =  -gV  bu.,  Ans, 

Rule.  Divide  the  fraction  of  the  lower  denomination  ly 
the  numbers  in  the  scale,  successively,  hetioee^i  the  given  and 
the  required  denomination. 

The  operation  will  frequently  be  shortened  by  cancellation. 

Examples  for  Practice. 

2.  What  part  of  a  rod  is  ^  of  a  foot  ?  Ans.  -^  rd. 

3.  What  part  of  a  pound  is  f  of  a  dram  ?      Ans.  y^^  lb. 

4.  Reduce  i^  of  a  cent  to  the  fraction  of  an  eagle. 

^ns.  ^-^  E. 
6.  A  hand  is  ^  of  a  foot ;  what  fraction  is  that  of  a  mile  ? 

6.  Reduce  |  of  2  pwt.  to  the  fraction  of  a  pound. 

Ans.  ^j^^  lb. 

7.  How  much  less  is  f  of  a  pint  than  -J-  of  a  hogshead  ? 

Ans.  IJf  hhd. 

8.  In  I  of  an  inch  what  fraction  of  a  mile  ? 

9.  f  of  an  ounce  Troy  is  f  of  what  fraction  of  2  pounds  ? 
10.  f  of  an  ounce  is  ^  of  what  fraction  of  2  pounds  Troy  ? 

Case  III. 
213.  To  reduce  a  denominate  fraction  to  integers 
of  lower  denominations. 
1.  What  is  the  value  of  |  of  a  hogshead  of  wine  ? 

Oive  explanation.     Rale.    Case  III  is  wliat  ? 

8* 


178  REDUCTION. 

OPERATION. 

I  hhd.  X  63=2^  gal.  =  39|  gal. 
I  gal.  X  4  =  J^  qt.  =  If  qt. ;  I  qt.  X  2  =  I  pt.  =  1  pt. 

Ans.  39  gal.  1  qt.  1  pt. 

Analysis,  f  hhd.  =  f  of  63  gal.,  or  39 1  gal. ;  and  f  gal.  =  f  of 
4  qt.,  or  1|  qt. ;  and  |  qt.  =  f  of  2  pt.,  or  1  pt. 

EuLE.  I.  Multiply  the  fraction  hy  that  numlerin  the  scale 
ivhich  will  reduce  it  to  the  next  lower  denomination,  and  if 
the  result  he  an  improper  fraction,  reduce  it  to  a  whole  or 
mixed  number. 

II.  Proceed  with  the  fractional  part,  if  any,  as  before, 
until  reduced  to  the  denominations  required. 

III.  The  units  of  the  several  denominations,  arranged  in 
their  order,  will  be  the  required  result. 

Examples  for  Practice. 

2.  Reduce  ^  of  a  month  to  lower  denominations. 

Ans.  17  da.  3  h.  25  min.  42^  sec. 

3.  What  is  the  value  of  f  of  a  £?     Ans.  8s.  6d.  3f  far. 

4.  What  is  the  value  of  -f  of  a  bushel  ? 

6.  Reduce  4  of  15  cwt.  to  its  equivalent  value. 

Ans.  12  cwt.  85  lbs.  llf  oz. 

6.  Reduce  f  of  f  of  a  pound  avoirdupois  to  integers. 

Ans.  4|4  oz. 

7.  What  is  the  value  of  \  of  an  acre  ?        Ans.  133^  P. 

8.  Reduce  4|  of  a  day  to  its  value  in  integers. 

Ans.  16  h.  36  min.  55^*^  sec. 

9.  What  is  the  value  of  f  of  a  pound  Troy  ? 

10.  What  is  the  value  of  -j-  of  5J  tons  ? 

Ans.  4  T.  5  cwt.  55J  lb. 

11.  What  is  the  value  of  |  of  3-|  a^jres  ?    Ans.l  A.  60  P. 

Case  IV. 
214.  To  reduce  a  compoiind  nTimber  to  a  fraction 
of  a  higher  denomination. 

1.  What  part  of  a  week  is  5  da.  14  h.  24  min.  ? 

Give  explanation.     Rule.     Case  VI  is  what  ? 


DENOMINATE    FRACTIONS.  179 

OPERATION.  Analysis.     To  find 

5  da.  14  h.  24  min.  =  8064  min.         what  part  one  compound 

1  Wk.  =  10080  min.  number  is   of    another, 

80  64    —  4  ^k     Ans  they  must  be  reduced  to 

TO 08  0         6        V  •  ^jjg  same  denomination. 

In  5  da.  14  h.  24  min.  there  are  8064  minutes,  and  in  1  week  there  are 

10080  minutes.    Since  1  minute  is  yjj^^^  of  a  week,  8064  minutes  is 

im%  =  I  of  a  week. 

Rule.  Reduce  the  given  number  to  its  lowest  denomination 
for  the  numerator,  and  a  unit  of  the  required  denomination 
to  the  same  denomination  for  the  denominator  of  the  required 
fraction. 

If  the  given  immber  contain  a  fraction,  the  denominator  of  this  fraction  must  be 
regarded  as  the  lowest  denomination. 

Examples  for  Practice. 

2.  What  part  of  a  mi.  is  266  rd.  3  yd.  2  ft.  ?      Ans.  I  mi. 

3.  What  fraction  of  a  £  is  13s.  7d.  3  far.  ? 

4.  Reduce  10  oz.  10  pwt.  10  gr.  to  the  fraction  of  a  pound 
Troy  ?  Ans.  \^  lb. 

5.  Reduce  2  cd.  ft.  8  cu.  ft.  to  the  fraction  of  a  cord. 

Ans.  -f^  Cd. 

6.  Reduce  1  bbl.  1  gal.  1  qt.  1  pt.  1  gi.  to  the  fraction  of 
a  hogshead.  Ans,  |-^  hhd. 

7.  What  part  of  2  rods  is  4  yards  1 J  feet  ?        Ans.  -^. 

8.  Reduce  l-f  pecks  to  the  fraction  of  a  bushel.    Ans.  f  bu. 

9.  What  part  of  9  feet  square  are  9  square  feet  Y 

10.  From  a  piece  of  cloth  containing  8  yd.  3  qr.  a  tailor 
cut  2  yd.  2  qr. ;  what  part  of  the  whole  piece  did  he  take  ? 

Ans.  f. 

Case  V. 

215.  To  reduce  a  denominate  decimal  to  integers 
of  lower  denominations. 

1.  Reduce  .78125  of  a  pound  Troy  to  integers  of  lower 
denominations. 

Give  explanation.     Rule.     Case  V  is  what  ? 


9.37500  oz. 
20 


180  REDUCTION'. 

OPERATION.  Analysis.     First  multiply  by  1 2 

.78125  lb.  to  reduce  the  given  number  from 

12  pounds  to  ounces,  and  the  result  is 

9  ounces  and  the  decimal  .375  of  an 

oz.     Then  multiply  this  decimal  by 

20  to  reduce  it  to  pennyweights, 

7.50000  pwt  ^^  S®*  "^  P^-  ^^^  .5  of  a  pwt. 

nA  This  last  decimal  we  multiply  by 

24,  to  reduce  it  to  grains,  and  the 

12.0000  gr.  result  is  12  gr.    Hence  the  answer 

9  OZ.  7  pwt.  12  gr.,  Ans,     is  9  oz.  7  pwt.  12  gr. 

EuLE.  I.  Miiltiply  the  given  decimal  hy  that  number  in  the 
scale  lohich  loill  reduce  it  to  the  next  lower  denomination,  and 
point  off  as  in  multiplication  of  decimals. 

II.  Proceed  with  the  decimal  part  of  the  product  in  the 
same  manner  until  reduced  to  the  required  denominations. 
The  integers  at  the  left  will  he  the  answer  required. 

Examples  for  Practice. 

2.  What  is  the  value  of  .217°  ?  Ans,  13'  1.2". 

3.  What  is  the  value  of  .659  of  a  week? 

Ans.  4  da.  14  h.  42  min.  43.2  sec. 

4.  Eeduce  .578125  of  a  bushel  to  integers  of  lower  denom- 
(nations.  Ans.  2  pk.  2  qt.  1  pt. 

5.  Reduce  .125  bbl.  to  integers  of  lower  denominations. 

Ans.  3  gal.  3  qt.  1  pt.  2  gi. 

6.  What  is  the  value  of  £.628125  ? 

7.  What  is  the  value  of  .22  of  a  hogshead  of  molasses  ? 

Ans.  13  gal.  3  qts.  3.52  gi. 
\  What  is  the  value  of  .67  of  a  league  ? 

Ans.  2  mi.  101  rd.  6  ft.  6.24  + in. 
^.  What  is  the  value  of  .42857  of  a  month  ? 

Ans.  12  da.  20  h.  34  min.  13^^  sec. 
IC.  What  is  the  value  of  .78875  of  a  long  ton  ? 

Ans.  15  cwt.  3  qr.  2  lb.  12.8  oz. 

Give  explanation.     Rule. 


DENOMINATE    FRACTIONS.  181 

11.  What  is  the  value  of  5.88125  acres  ?   A^is,  5  A.  141  P. 

12.  Eeduce  .0055  T.  to  pounds.  A7is,  11  lb. 

13.  Reduce  .034375  of  a  bundle  of  paper  to  its  value  in 
lower  denominations.  Ans.  1  quire  9  sheets. 

Case  VI. 

216.  To  reduce  a  compound  number  to  a  decimal 
of  a  higher  denomination. 

1.  Reduce  3  pk.  2  qt.  to  the  decimal  of  a  busheL 

OPERATION.  Analysis.     Since  8  quarts  make 

2.00       qt.  1  peck,  and  4  pecks  1  bushel,  there 

o  .^Krvrj     1^  "^1^  ^6  I  as  many  pecks  as  quarts 

— ^   '  (183),  and  I  as  many  bushels  as 

.8125  bu.,  Ans.  pecks. 

Or,  3  pk.  2  qt.  =  26  qt.  ^"^  ^'^  ^^^  ^^^^^^  ^  P^-  ^  qt-  to 

^\^                Qo     f  *^^  fraction  of  a  bushel  (as  in  214), 

^  ><  ^^^  ^^  h&ve  f  I  of  a  bushel,  which, 

|-|  =  .8125  bu.,  Ans.  reduced  to  a  decimal,  equals  .8125. 

Rule.  Divide  the  lowest  denomination  given  by  that  num- 
ber in  the  scale  which  will  reduce  it  to  the  next  higher,  and 
annex  the  quotient  as  a  decimal  to  that  higher.  Proceed  in 
the  same  manner  until  the  whole  is  reduced  to  the  denomina- 
tion required.     Or, 

Reduce  the  given  number  to  a  fraction  of  the  required  de- 
nomination, and  reduce  this  fraction  to  a  decimal. 

Examples  foe  Pkactice. 

2.  Reduce  3  qt.  1  pt.  1  gi.  to  the  decimal  of  a  gallon. 

Ans.  .90625  gal. 

3.  Reduce  10  oz.  13  pwt.  9  gr.  to  the  decimal  of  a  pound 
Troy.  Ans.  .8890625  lb. 

4.  Reduce  1.2  pints  to  the  decimal  of  a  hogshead. 

^?25.  .00238+  hhd. 

5.  What  part  of  a  bushel  is  3  pk.  1.12  qt.  ?    Ans.  .785  bu. 

Case  VI  is  what?    Give  explanations.     Rule. 


182  ADDITION. 

6.  What  part  of  an  acre  is  132.56  P.  ? 

7.  Reduce  17  yd.  1  ft.  6  in.  to  the  decimal  of  a  mile. 

Ans.  .00994318+  mi. 

8.  Reduce  .32  of  a  pint  to  the  decimal  of  a  bushel. 

Ans.  .005  bu. 

9.  Reduce  4-J  feet  to  the  decimal  of  a  fathom. 

Ans.  .8125  fathom, 

10.  Reduce  150  sheets  of  paper  to  the  decimal  of  a  ream. 

Ans.  .3125  Rm. 

11.  Reduce  47.04  lb.  of  flour  to  the  decimal  of  a  barrel. 

12.  Reduce  .33  of  a  foot  to  the  decimal  of  a  mile. 

13.  Reduce  5  h.  36  min.  57^^  sec.  to  the  decimal  of  a  day. 


ADDITION. 

217.  1.  A  miner  sold  at  one  time  10  lb.  4  oz.  16  pwt. 
8  gr.  of  gold;  at  another  time,  2  lb.  9  oz.  3  pwj.;  at  another, 
11  oz.  20  gr. ;  and  at  another,  25  lb.  16  pwt.  23  gr. ;  how 
much  did  he  sell  in  all  ? 

Analysis.  Arranging  the  num- 
bers in  columns,  placing  units  of  the 
same  denomination  under  each  other, 
we  first  add  the  units  in  the  right 
hand  column,  or  lowest  denomina- 
tion, and  find  the  amount  to  be  51 
grains,  which  is  equal  to  2  pwt. 
Ans.  Sd        1      17       3  3  gr.     We  write  the  3  gr.  under  the 

column  of  grains,  and  add  the  2  pwt. 
to  the  column  of  pwt.  We  find  the  amount  of  the  second  column  to 
be  37  pwt.,  which  is  equal  to  1  oz.  17  pwt.  Writing  the  17  pwt.  under 
the  column  of  pwt.,  we  add  the  1  oz.  to  the  next  column.  Adding  this 
column  in  the  same  manner  as  the  preceding  ones,  we  find  the  amount 
to  be  25  oz,,  equal  to  2  lb.  1  oz.  Placing  the  1  oz.  under  the  column 
of  oz.,  we  add  the  2  lb.  to  the  column  of  lb.  Adding  the  last  column, 
we  find  the  amount  to  bo  39  lb. 


OPERATION. 

lb.      oz.   pwt. 

gr. 

10       4     16 

8 

2       9       3 

0 

0     11       0 

20 

25       0     16 

23 

What  is  addition  of  compound  numbers  ?    Give  exi)lanation. 


COMPOUND    liTUMBERS.  183 

EuLE.  I.  Write  the  numbers  so  that  those  of  the  same  unit 
value  will  stand  in  the  same  column. 

II.  Beginning  at  the  right  hand,  add  each  denomination  as 
in  simple  numbers,  carrying  to  each  succeeding  denomination 
one  for  as  many  units  as  it  tahes  of  the  denomination  added, 
to  mahe  one  of  the  next  higher  denomination. 

Examples  for  Practice. 


(2.) 

< 

:3.) 

£.   s.   d. 

fe 

I 

3   3  gr. 

48  13   8 

12 

8 

7  2  15 

51   6   4 

10 

4  1  10 

67  11   3 

15 

0 

2  1  19 

76  18  10 

11 

6  0  12 

244  10   1 

13 

4 

4  2   0 

(4.) 

(5.) 

T.  cwt.  lb.  i 

DZ. 

bn. 

pk.  qt.  pt. 

4   7  18 

4 

1 

3  7  1 

15  98 

15 

3 

2  2  0 

3   9  10 

6 

16  1 

1   0  15 

0 

17 

0  5  1 

9  12  42 

9 

45 

2  4  0 

6.  What  is  the  sum  of  4  mi.  150  rd.  2  yd.  1  ft.  10  in., 
5  mi.  258  rd.  1  yd.  2  ft.  6  in.,  10  mi.  185  rd.  2  yd.  2  ft. 
11  in.,  and  268  rd.  4  yd.  2  ft.  1  in.  ? 

7.  Find  the  sum  of  197  sq.  yd.  4  sq.  ft.  104|  sq.  in.,  122 
sq.  yd.  2  sq.  ft.  27}  sq.  in.,  5  sq.  yd.  8  sq.  ft.  2|  sq.  in.,  and 
237  sq.  yd.  7  sq.  ft.  128-^  sq.  in.  ? 

Ans,  563  sq.  yd.  4  sq.  ft.  118.825  sq.  in. 

When  common  fractions  occur,  they  should  be  reduced  to  a  common  denomina- 
tor, to  decimals,  or  to  integers  of  a  lower  denomination,  and  added  according  to  the 
usual  method. 

Give  tbe  Rule. 


184 


ADDITIOI?^. 

A. 

26 

P. 

148 

(8.) 

sq  yd.  1 
25 

sq.  ft.     sq.  In. 
8         125 

19 

118 

30 

7        150 

456 

100 

16 

6          98 

603 

48 

5          85 

■-Hi) 

(i)   =  72 

503 

48 

13 

1          13 

mi. 
1 

(9.) 
rd.    yd. 
310     4 

ft.    in. 
2     11 

(10.) 
hhd.    gal.    qt  pt 
27     65     3     2 

3 

160     2 

1     10 

112    60    2    3 

10 

305     1 

2     11 

60    29    0    1 

16 

136     3i 

1      8 

421      0    2    3 
14    39     1    2 

bu. 
23 

(11.) 

pk.    qt. 

3       7 

pt. 

1 

yr. 
25 

(12.) 
da.     hr.   min.  sec. 
300     19     54    35 

34 

2      0 

1 

21 

40     12    40     24 

42 

3      5 

0 

3 

112     14    15     17 

61 

1      4 

1 

6 

19     11     45     69 

23 

0      3 

0 

1 

1111 

11 

3      4 

0 

57 

109     11     37     16 

13.  If  a  printer  one  day  use  4  bundles  1  ream  15  quires 
20  sheets  of  paper,  the  next  day  3  bundles  1  ream  10  quires 
10  sheets,  and  the  next  2  bundles  13  sheets,  how  much  does 
he  use  in  the  three  days  ? 

Ans.  2  bundles  1  ream  6  quires  19  sheets. 

14.  A  tailor  used,  in  one  year,  2  gross  5  doz.  10  buttons, 
another  year  3  gross  7  doz.  9,  and  another  year  4  gross  6 
doz.  11 ;  how  many  did  he  use  in  the  three  years  ? 

Ans.  10  gross  8  doz.  6. 


i 


COMPOUND    NUMBERS.  185 

15.  A  ship,  leaving  New  York,  sailed  easi  the  first  day 
3°  45'  50" ;  the  second  day,  4°  50'  10"  ;  the  third,  2°  10'  55"; 
the  fourth,  2°  39"  ;  how  far  was  she  then  east  from  the  place 
of  starting  ?  Ans.  12°  47'  34 ". 

16.  A  man,  in  digging  a  cellar,  removed  127  cu.  yd. 
20  cu.  ft.  of  earth ;  in  digging  a  drain,  6  cu.  yd.  25  cu.  ft. ; 
and  in  digging  a  cistern,  17  cu.  yd.  18  cu.  f  t. ;  what  was 
the  amount  of  earth  removed,  and  what  the  cost  at  1 6  cents 
a  cu.  yd.  ?  Ans,  152|  cu.  yd. ;  $24.37^. 

17.  A  farmer  received  80  ce»ts  a  bushel  for  4  loads  of 
com,  weighing  as  follows  :  2564,  2713,  3000,  and  3109  lbs.; 
how  much  did  he  receive  for  the  whole  ?    Ans.  $162,657. 

18.  A  druggist  sold  for  medicine,  in  three  years,  at  an  aver- 
age price  of  9  cents  a  gill,  the  following  amounts  of  brandy, 
viz.  :  1  bbl.  4  gal.  1  pt. ;  30  gal.  2  qt.  1  gi. ;  2  bbl.  15  gal. ; 
how  much  did  he  receive  for  the  whole  ?       Ans,  $415.17. 

218.  To  add  denominate  fractions. 

1.  Add  f  of  a  mile  to  13^  rods. 

OPERATION.  Analysis.    Find  the  value 

4  mi.  =  266  rd.  11     ft.  '           ^^  ^^^^  fraction  in  integers 

mrd.'=    13  rd.'    Sift!  of  less  denominations  (213). 

^               ± —  and  then  add  tneir  values  as 

.^715.  280  rd.    0  in  compound  numbers  (217). 

Or,  13i  rd.-^320=^V  mi.  ^^'  ^^^^^  *^^  ^^^°  ^^^^- 

,        -.K       •       91       •        nor\    A  tions  to  fractions  of  the  same 

A  mi.  + 1- mi. =14  mi.  =280  rd.  .         .    ^.       ,«-,ox    ^. 

2*           '   ^              2*  denomination    (213),   the^ 

add  them,  and  find  the  value  of  their  sum  in  lower  denominations  (213) 

2.  Add  J  of  a  rod  to  }  of  a  foot.  Ans,  13  ft.  1^  in. 

3.  What  is  the  sum  of  -J  of  a  mile,  28^  rods  ? 

Ans,  308  rd.  2  ft.  9  in. 

4.  What  is  the  sum  of  f  of  a  pound  and  -J  of  a  shilling  ? 

Ans.  13s.  lOd.  2f  qr. 

5.  What  is  the  sum  of  |  of  a  ton  and  ^  of  1  cwt.  ? 

Ans.  12  cwt.  42  lb.  134  oz. 

Give  explanation  of  the  process  of  adding  denominate  fractions- 


186  SUBTRACTION. 

6.  What  is  the  sum  of  |  of  a  day  added  to  J  an  hour  ? 

Ans.  9  h.  30  min. 

7.  What  is  the  sum  of  |  of  a  week,  J  of  a  day,  and  ^  of 
an  hour?  •  Ans,  1  da.  22  h.  15  min. 

8.  Add  -f  of  a  hhd.  to  }  of  a  gal. 

9.  What  is  the  sum  of  4^  of  a  cwt.,  8|-  lb.,  and  3^  oz.  by 
long  ton  table  ?  Ans,  73  lb.  1-^  oz. 

10.  What  is  the  sum  of  |  of  a  mile,  f  of  a  yard,  and  J  of 
afoot? 

11.  Sold  4  village  lots;  the  first  contained  J  of  ^  of  an 
acre ;  the  second,  60J  rods ;  the  third,  -f-  of  an  acre ;  and 
the  fourth,  |  of  |  of  an  acre  ;  how  much  land  in  the  four 
lots  ?  Ans.  146  P.  126^  sq.  ft. 

12.  A  farmer  sold  three  loads  of  hay ;  the  first  weighed 
1^  T.,  the  second,  1^  T.,  and  the  third,  18|-  cwt. ;  what 
was  the  aggregate  weight  of  the  three  loads  ? 

Am.  3  T.  5  cwt.  91  lb.  10|  oz. 

SUBTRACTION. 

319.  1.  If  a  druggist  buy  25  gal.  2  qt.  1  pt.  1  gi.  of  wine, 
and  sell  18  gal.  3  qt.  1  pt.  2  gi.,  how  much  has  he  left  ? 

Analysis.  Writing  tlie  subtrahend 
under  the  minuend,  placing  units  of  the 
same  denomination  under  each  other, 
we  begin  at  the  right  hand,  or  lowest 
denomination ;  since  we  cannot  take 
2  gi.  from  1  gi.,  we  add  1  pt.  or  4  gi.  to 
i  gi. ,  making  5  gi. ;  and  taking  2  gi.  from  5  gi.,  we  write  the  remain- 
der, 3  gi.,  underneath  the  column  of  gills.  Having  added  1  pt.  or 
4  gi.  to  the  minuend,  we  now  add  1  pt.  to  the  0  pt.  in  the  subtra- 
hend, making  1  pt. ;  and  1  pt.  from  1  pt.  leaves  0  pt.,  which  we  write 
in  the  remainder.  Next,  as  v»e  cannot  take  3  qt.  from  2  qt.,  we  add 
1  gal.  or  4  qt.  to  2  qt.,  making  6  qt.,  and  taking  3  qt.  from  6  qt.,  we 
write  the  remainder,  3  qt.,  under  the  denomination  of  quarts.  Add- 
ing 1  gal.  to  18  gal.,  we  subtract  19  gal.  from  25  gal.,  as  in  simple 


OPERATION. 

gal.     qt.    pt. 
25      2      1 
18      3      0 

gi. 
1 
2 

Ans.    6      3      0 

3 

What  is  subtraction  of  compound  numbers  ?    Give  explanation. 


COMPOUI^^D     K  UMBERS.  187 

numbers,  and  write  the  remainder,  6  gal.,  under  the  column  of  gal- 
lons. 

KuLE.  I.  Write  the  subtrahend  U7ider  the  minuend,  so  that 
units  of  the  same  denomination  shall  stand  under  each  other. 

II.  Beginning  at  the  right  hand,  subtract  each  denomina- 
tion separately,  as  in  simple  numbers. 

III.  If  the  number  of  any  denomination  in  the  suhtrahend 
exceed  that  of  the  same  denomination  in  the  miriuend,  add  to 
the  number  in  the  minuend  as  many  units  as  make  one  of  the 
next  higher  denomination,  and  then  subtract;  in  this  case 
add  1  to  the  next  higher  denomination  of  the  suMrahend 
before  subtracting.  Proceed  m  the  same  maimer  with  each 
denomination. 

Examples  for  Practice. 


(2. 

) 

(3.) 

lb. 

oz. 

pwt. 

gr- 

A. 

P. 

From  18 

6 

10 

14 

25 

96.9 

Take  10 

5 

4 

6 

19 

145.14 

Rem.   8 

1 

6 

8 

5 

111.76 

(4.) 

(5.) 

T.   cwt. 

lb. 

yr. 

da. 

h. 

min.  Bee. 

14  n 

69J 

38 

187 

16 

45   50 

10   12 

98| 

17 

190 

20 

50   40 

20      361       19       55       10 

6.  A  Boston  merchant  bought  English  goods  to  the 
amount  of  £4327  13s.  7id.,  and  he  paid  £1374  10s.  llfd. ; 
how  much  did  he  then  owe  ? 

7.  From  300  miles  take  198  mi.  305  rd.  2  yd.  1  ft.  10  in. 

Ans.  101  mi.  14  rd.  2  yd.  2  ft.  8  in. 

8.  "What  is  the  difference  in  the  longitude  of  two  places, 
one  75°  20'  30"  west,  and  the  other  71°  19'  35"  west? 

Ans.  4°  55". 

9.  From  lOib  7  1  4  3  1 3  15  gr.  take  3ib  8  §  2  3  23  18  gr. 

Ans.  6ft)  11  §  13  13  17gr. 

Give  the  Rule. 


188  SUBTRACTION. 

10.  The  apparent  periodic  revolution  of  the  sun  is  made 
in  365  da.  6  h.  9  min.  9  sec,  and  that  of  the  moon  in  29  da. 
12  h.  44  min.  3  sec. ;  what  is  the  difference  ? 

Ans,  335  da.  17  h.  25  min.  6  sec. 

11.  A  man,  having  a  hogshead  of  wine,  drank,  on  an  aver- 
age, for  five  years,  including  two  leap  years,  one  gill  of  wine 
a  day  ;  how  much  remained  ?      A^is.  5  gal.  3  qt.  1  pt.  1  gi. 

12.  A  section  of  land  containing  640  acres  is  owned  by 
four  men  ;  the  first  owns  196  A.  96 J  P. ;  the  second,  200  A. 
60  P. ;  the  third,  177  A.  36  P. ;  how  much  does  the  fourth 
own  ?  Ans.  65  A.  127.75  P. 

13.  From  a  pile  of  wood  containing  75f  Cd.  was  sold  at 
one  time  16  Cd.  5  cd.  ft. ;  at  another,  24  Cd.  6  cd.  ft.  12 
cu.  ft. ;  at  another,  27  Cd.  112  cu.  ft.  ;  how  much  remained 
in  the  pile  ?  Aiis.  6  Cd.  3  cd.  ft.  4  cu.  ft. 

14.  If  from  a  hogshead  of  molasses  10  gal.  1  qt.  1  pt.  be 
drawn  at  one  time,  15  gal.  1  pt.  at  another,  and  14  gal.  3  qt. 
at  another,  how  much  will  remain  ? 

230.  To  find  the  difference  in  dates. 

1.  What  length  of  time  elapsed  from  the  discovery  of 
America  by  Columbus,  Oct.  14,  1492,  to  the  Declaration  of 
Independence,  July  4,  1776? 

FIRST  OPERATION.  ANALYSIS.    Place  the  earlier  date  un- 

yr.  mo.       da.  der  the  later,  writing  first  on  the  left 

1776  7  4  the  number  of  the  year  from  the  Chris- 

1492        10        14  ^^^^  6ra,  next  the  number  of  the  month, 

'  ~r~  7        TT  counting  January  as  the  first  month,  and 

yioo  o       <^\}  j^g^j.  ^i^g  number  of  the  day  from  the 

first  day  of  the  month.    Instead  of  the  number  of  the  year,  month, 

and  day,  some  use  the  number  of  years,  months,  and  days  that  ham 

elapsed  since  the  Christian  era,  thus : 
instead  of  saying  July  is  the  7th  mouth, 
we  say  6  months  and  3  days  have 
elapsed,  and  instead  of  saying  October 
is  the  10th  month,  we  say  9  months  and 
283  8        20  13  days  have  elapsed. 

How  is  the  difference  of  dates  found  ? 


SECOND 

OPERATION. 

yr. 

mo.       da. 

1775 

6         3 

1491 

9       13 

COMPOUlN^D     NUMBEKS. 


189 


Both  methods  will  obtain  the  same  result ;  the  former  is  generally 
used. 

1.  When  hours  are  to  be  obtained,  we  reckon  from  12  at  night,  and  if  minutes  and 
seconds,  we  write  them  still  at  the  right  of  hours. 

2.  In  finding  the  time  between  two  dates,  or  in  computing  interest,  12  months  are 
considered  a  year,  and  30  days  a  month. 

When  the  exact  number  of  days  is  required  for  any  period 
not  exceeding  one  ordinary  year,  it  may  be  readily  found  by 
the  following 

Table, 

Shomng  the  number  of  days  from  any  day  of  one  month  to  the  same  day 

of  any  other  month  within  one  year. 


raOM  ANT  DAT 

TO  THE  SAJfE  DAT  OP  THiS  NBXT. 

or 

Jan. 
365 

Feb. 
31 

59 

Apr. 
90 

May 

Jane 

July 

Aug. 

Sept. 

Oct. 

Nov. 

Dec. 

January 

120 

151 

181 

212 

243 

273 

304 

334 

February  

a34 

365 

28 

59 

89 

120 

150 

181 

212 

242 

273 

303 

March 

306 

337 

365 

31 

61 

92 

122 

153 

1&4 

214 

245 

275 

April 

275 

306 

334 

365 

30 

61 

91 

122 

153 

183 

214 

244 

May 

245 

276 

304 

335 

365 

31 

61 

92 

123 

153 

184 

214 

J  una      

214 

184 

245 
215 

273 
243 

304 
274 

as4 

304 

365 
335 

30 
365 

61 
31 

92 
62 

122 
92 

153 
123 

183 

July 

153 

August 

153 

184 

212 

243 

273 

304 

334 

365 

31 

61 

92 

122 

September 

122 

153 

181 

212 

242 

273 

303 

334 

365 

30 

61 

91 

October 

92 

123 

151 

182 

212 

243 

273 

304 

335 

365 

31 

61 

November 

61 

92 

120 

151 

181 

212 

242 

273 

304 

334 

365 

30 

December 

31 

62 

90 

121 

151 

182 

212 

243 

274 

304 

335 

365 

If  the  days  of  the  different  months  are  not  the  same,  the 
number  of  days  of  difference  should  be  added  when  the 
earher  day  belongs  to  the  month /row  which  we  reckon,  and 
siibtracted  when  it  belongs  to  the  month  to  which  we  find 
the  time.  If  the  29th  of  February  is  to  be  included  in  the 
time  computed,  one  day  must  be  added  to  the  result. 

Examples  foe  Practice. 
2.  George  Washington  was  bom  Feb.  22,  1732,  and  died 
Dec.  14,  1799  ;  what  was  his  age  ? 

Arts.  67  yr.  9  mo.  22  da. 

How  can  the  number  of  days,  if  less  than  a  year,  be  obtained  ? 


IDO  SUBTRACTION". 

3.  How  much  time  has  elapsed  since  the  declaration  of 
independence  of  the  United  States? 

4.  How  many  years,  months,  and  days  from  your  birth- 
day to  this  date  ;  or  what  is  your  age  ? 

5.  How  long  from  the  battle  of  Bunker  Hill,  June  17, 1775, 
to  the  battle  of  Waterloo,  June  18, 1815  ?    Ans.  40  yr.  1  da. 

6.  What  length  of  time  will  elapse  from  20  minutes  past 
2  o'clock,  p.  M.,  June  24, 1856,  to  10  minutes  before  9  o'clock, 
A.  M.,  January  3,  1861  ?     Ans.  4  yr.  6  mo.  8  da.  18  h.  30  min. 

7.  How  many  days  from  any  day  of  April  to  the  same  day 
of  August  ?  of  December  ?  of  February  ? 

8.  How  many  days  from  the  6th  of  November  to  the  15th 
of  April  ?  Ans.  160  days. 

9.  How  many  days  from  the  20th  of  August  to  the  15th 
of  the  following  June  ?  Ans.  299  days. 

231.  To  subtract  denominate  firactions. 

1.  From  f  of  an  oz.  take  J  of  a  pwt. 

OPEKATiON.  Analysis.    We  per- 

■|  oz.      =7  pwt.  12  gr.  form  the  same  reduc- 

<»  ^^^^  __  21  ST.  tions  as  in  addition  of 

denominate    fractions, 

6  pwt.  15  gr.,  Ans.  ( 2 1 8 ),  and  then  sub- 

„  o  «^        »„        I  tract    the    less    value 

Or,      |oz.x20  =  i^^pwt.  from  the  greater, 
iyi-  —  -J  =  -5/  pwt.=  6  pwt.  15  gr. 

2.  What  is  the  difference  between  ^  rod  and  |  of  a  foot  ? 

Ans.  7  ft.  6  in. 

3.  From  £f  take  |  of  f  of  a  shilling. 

4.  From  f  of  a  league  take  i^  of  a  mile. 

Ans.  1  mi.  193  rd.  11  ft.  7.9  in.+ 

5.  From  8^2^^  cwt.  take  27h}  lb. 

Ans.  8  cwt.  62  lb.  9|  oz. 

6.  From  -^  of  a  week  take  ^  of  a  day. 

A  ns.  1  da.  4  h.  48  min. 

Give  explanation  of  the  process  of  subtracting  denominate  fractions. 


C0MP0UKD:N^  UMBERS.  191 

7.  Two  persons,  A  and  B,  start  from  two  places  120  miles 
apart,  and  travel  toward  each  other  ;  after  A  travels  f ,  and 
B  f,  of  the  distance,  how  far  are  they  apart? 

Ans.  41  mi.  289  rd.  8  ft.  7^  in. 

8.  From  a  cask  of  brandy  containing  96  gallons,  \  leaked 
out,  and  f  of  the  remainder  was  sold ;  how  much  still  re- 
mained in  the  cask  ?  Ans.  25  gaL  2  qt.  3-J  gi. 


MULTIPLICATION. 

222,  1.  A  farmer  has  8  fields,  each  containing  4  A. 
107  P. ;  how  much  land  in  all? 

OPEBATION.  Analysis.    In  8  fields  are  8  times  as  much 

A.        P.  land  as  in  1  field.     We  write  the  multiplier 

4      107  under  the  lowest  denomination  of  the  multi- 

8  plicand,  and  proceed  thus :  8  times  107  P. 

— -~  are  856  P.,   equal  to  5  A.  56  P.  ;   and  we 

'^  *         ^^  write  the  56  P.  under  the  number  multiplied. 

Again,  8  times  4  A.  are  32  A.,  and  5  A.  added  make  87  A.,  which  we 
write  under  the  same  denomination  in  the  multiplicand,  and  the  work 
is  done. 

Rule.  I.  Write  the  multiplier  under  the  lowest  denomina" 
tion  of  the  multiplicand. 

II.  Multiply  as  in  simple  numbers,  and  carry  as  in  addi- 
tion  of  compound  numbers. 


Examples 

FOR 

Practice. 

(3.) 

bu.     pk.     qt.     pt. 

4       2       5       1 

2 

(3.) 
mi       rd. 
9       180 

ft. 

13 

6 

9       13       0 

57      124 

12 

Multiplication  of  compound  numbers,  how  perfonned  ?    Rule. 


192 


MULTIPLICATION. 


(4.) 

(5.) 

£ 

s.        d. 

lb.     oz.    pwt.    gr. 

5 

18       4 

3       4       0       22 

4 

7 

(6.) 

(7.) 

T. 

cwt.       lb. 

oz. 

14 

16       48 

12 

13°       10'      35" 

11 

9 

8.  In  6  barrels  of  grain,  each  containing  2  bu.  3  pk.  5  qt., 
how  many  bushels?  Ans.  17  bu.  1  pk.  6  qt. 

9.  If  a  druggist  deal  out  31b4§  I3  23l6gr.  of  medicine 
a  day,  how  much  will  he  deal  out  in  6  days  ? 

10.  If  a  man  travel  29  mi.  150  rd.  15  ft.  in  1  day,  how 
far  will  he  travel  in  8  days  ? 

11.  If  a  woodchopper  can  cut  3  Od.  48  cu.  ft.  of  wood  in  1 
day,  how  many  cords  can  he  cut  in  12  days  ?    Ans.  40^  Cd. 

12.  What  is  the  weight  of  48  loads  of  hay,  each  weighing 
IT.  3  cwt.  50  1b.? 

Analysis.  When  the  multi- 
plier is  large,  and  a  composite 
number,  we  may  multiply  by  one 
of  the  factors,  and  that  product 
by  the  other.  Multiplying  the 
weight  of  1  load  by  6,  we  obtain 
the  weight  of  6  loads,  and  the 
weight  of  6  loads  multiplied  by 
8,  gives  the  weight  of  48  loads. 

13.  If  1  acre  of  land  produce  45  bu.  3  pk,  6  qt.  1  pi  of 
com,  how  much  will  64  acres  produce?       Ans.  2941  bu. 

14.  How  much  will  120  yards  of  cloth  cost,  at  £1  9s.  8}d. 
per  yard  ? 

15.  If  $80  will  buy  4  A.  146  P.  20  sq.  yd.  3  sq.  ft.  of  land, 
how  much  will  $4800  buy?  Ans.  295  A.  10  sq.  yd. 

16.  If  a  load  of  coal  by  the  long  ton  weigh  1  T.  6  cwt.  2  qr. 
26  lb.  10  oz.,  what  will  be  the  weight  of  73  loads  ? 

Ans.  97  T.  11  cwt.  3  qr.  11  lb.  10  oz. 


OPERATION. 
T.     cwt.      lb. 

1      3       50 
6 


00 

8 


weight  of  6  loads. 


56        8        00  weight  of  48  loads. 


COMPOUND    NUMBERS.  193 

17.  The  sun,  on  an  average,  changes  his  longitude  59' 
8.33"  per  day  ;  how  much  will  be  the  change  in  365  days  ? 

18.  If  1  pt.  3  gi.  of  wine  fill  1  bottle,  how  much  will  be 
required  to  fill  a  great  gross  of  bottles  of  the  same  capacity  ? 

DIVISION. 

223.  1.  If  4  acres  of  land  produce  102  bu.  3  pk.  2  qt.  of 

wheat,  how  much  will  1  acre  produce  ? 

OPEKATiON.  Analysis.    One  acre  will  produce  { 

bu.    pk.     qt.     pts.  as  much  as  4  acres.     Writing  the  divi- 

4  )  102     3       2  (5or  on  the  left  of  the  dividend,  we  di- 

~  ~  ~  7~  vide  103  bu.  by  4,  and  obtain  a  quotient 
of  35  bu.,  and  a  remainder  of  3  bu.  We 
write  the  25  bu.  under  the  denomination  of  bushels,  and  reduce  the 
3  bu.  to  pecks,  making  8  pk.,  and  the  3  pk.  of  the  dividend  added 
makes  11  pk.  Dividing  11  pk,  by  4,  we  obtain  a  quotient  of  3  pk. 
and  a  remainder  of  3  pk. ;  writing  the  3  pk.  under  the  order  of  pecks, 
we  next  reduce  3  pk.  to  quarts,  adding  the  3  qt.  of  the  dividend,  mak- 
ing 36  qt.,  which  divided  by  4  gives  a  quotient  of  6  qt.  and  a  remainder 
of  3  qt.  Writing  the  6  qt.  under  the  order  of  quarts,  and  reducing  the 
remainder,  3  qt. ,  to  pints,  we  have  4  pt. ,  which  divided  by  4  gives  a 
quotient  of  1  pt. ,  which  we  write  under  the  order  of  pints,  and  the 
work  is  dona 

2.  A  farmer  put  132  bu.  operation. 

1  pk.  of  apples  into  46  bar-  ^^*    P^- 

rels  ;  how  many  bu.  did  he     ^^)^^^    1  ( ^  bu, 
put  into  a  barrel  ?  _ 

40 
4 

When  the  divisor  is  large,  and  161  (  3  pk. 

not  a  composite  number,  we  di-  138 

vide  by  long  division,  as  shown  oQ 

in  the    operation.      From    these 
examples  we  derive  the 


8 


184  (  4  qt. 

—  Ans.  2  hu.S])k.  4.  qt 


Explain  the  process  of  dividing  compound  numbers. 
R.  P.  0 


194  DIVISION. 

Rule.  I.  Divide  the  highest  denomination  as  in  simple 
numbers^  and  each  succeeding  denomination  in  the  same 
manner,  if  there  he  no  remainder, 

II.  If  there  he  a  remainder  after  dividing  any  denomina- 
tion, reduce  it  to  the  next  lower  denomination,  adding  in  the 
given  number  of  that  denomination,  if  any,  and  divide  as  before, 

III.  The  several  partial  quotients  will  he  the  quotient  re- 
quired. 

1.  When  the  divisor  is  large  and  is  a  composite  number,  we  may  shorten  the  work 
by  dividing  by  the  factors. 

2.  When  the  divisor  and  dividend  are  both  compound  numbers,  they  must  both 
be  reduced  to  the  same  denomination  before  dividing,  and  then  the  procesB  is  the 
same  as  in  simple  numbers. 


Examples 

FOE 

Practice. 

(3.) 

(4.) 

£ 

s. 

d. 

T. 

cwt. 

lb. 

5)25 

8 

4 

7)45 

15 

25 

5 

1 

8 

6 

10 

75 

(5.) 

wk.    da. 

h. 

TTiin. 

(6.) 

4)3       5 

22 

00 

10)25° 

42' 

40" 

6 

17 

30 

2 

34 

16 

7.  Bought  6  large  silver  spoons,  which  weighed  11  oz. 
3  pwt. ;  what  was  the  weight  of  each  spoon  ? 

8.  A  man  traveled  by  railroad  1000  miles  in  one  day; 
what  was  the  average  rate  per  hour  ? 

Ans.  41  mi.  213  rd.  5  ft.  6  in. 

9.  If  a  family  use  10  bbl.  of  flour  in  a  year,  what  is  the 
average  amount  each  day?  Ans.  5  lb.  5fJ  oz. 

10.  The  aggregate  weight  of  123  hogsheads  of  sugar  is 
57  T.  19  cwt.  42  lb.  14  oz.;  what  is  the  average  weight  per 
hogshead  ?  Ans.  9  cwt.  42  lb.  10  oz. 

11.  How  many  times  are  £5  10s.  lOd.  contained  in  £537 
10s.  lOd.  ?  Ans.  97. 

Give  the  rule.  Wlien  the  divisor  is  a  composite  number,  how  may 
we  proceed  ?  When  the  divisor  and  dividend  are  both  compound  num- 
bers,  how  proceed  ? 


COMPOCND    NUMBERS.  195 

12.  A  cellar  50  ft.  long,  30  ft.  wide,  and  6  ft.  deep  was 
excavated  by  5  men  in  6  days ;  how  many  cubic  yards  did 
each  man  excavate  daily?  Ans.  11  cu.  yd.  3  cu.  ft. 

13.  If  a  town  5  miles  square  be  divided  equally  into  150 
farms,  what  will  be  the  size  of  each  farm  ? 

Ans.  106  A.  106  P.  20  sq.  yd.  1  sq.  ft.  72  sq.  in. 

14.  How  many  times  are  4  bu.  3  pk.  2  qt.  contained  in 
336  bu.  3  pk.  4  qt.  ?  Ans.  70  times. 

15.  A  merchant  tailor  bought  4  pieces  of  cloth,  each  con- 
taining 60  yd.  2.25  qr. ;  after  selling  J-  of  the  whole,  he 
made  up  the  remainder  into  suits  containing  9  yd.  2  qr. 
each  ;  how  many  suits  did  he  make  ?  A7is.  17. 


LONGITUDE    AND    TIME. 

234.  Every  circle  is  supposed  to  be  divided  into  360 
equal  parts,  called  degrees. 

Since  the  sun  appears  to  pass  from  east  to  west  round  the 
earth,  or  through  360°,  once  in  every  24  hours,  it  will  pass 
through  ^  of  360°,  or  15°  of  the  distance,  in  1  hour  ;  ana 
1°  of  distance  in  ^i^  of  1  hour,  or  4  minutes ;  and  1'  of 
distance  in  -^  of  4  minutes,  or  4  seconds. 

Table  of  Longitude  and  Time. 

360°  of  longitude  =  24  hours,  or  1  day  of  time. 

15°  *'        "  =1  hour  '*      " 

1°  ''        "         =    4  minutes  "      " 

V   "        "         =    4  seconds  **     " 

Case  I. 

225.  To  find  the  diflference  of  time  between  two 
places,  when  their  longitudes  are  given. 

1.  The  longitude  of  Boston  is  71°  3',  and  of  Chicago  87° 
30' ;  what  is  the  difference  of  time  between  these  two  places  ? 

Explain  how  distance  is  measured  by  time.  Repeat  the  table  ot 
lon'^itude  and  time.     Case  I  is  what? 


OPERATION. 

87° 

30' 

71° 

3' 

16° 

27' 

4 

196  LONGITUDE    AKD    TIME. 

Analysis.  By  subtrRction  of 
compound  numbers  we  first  find 
the  difllerence  of  longitude  be- 
tween the  two  places,  which  is 
16°  27'.  Since  1°  of  longitude 
makes  a  difference  of  4  minutes 

1  h.  5  min.  48  sec,  Ans.  ^^  ^^^^>  ^^^  1'  ^^  longitude  a 

difference  of  4  seconds  of  time, 
we  multiply  16°  27',  the  difference  in  longitude,  by  4,  and  we  obtain 
the  difference  of  time  in  minutes  and  seconds,  which,  reduced  to  higher 
denominations,  gives  1  h.  5  min.  48  sec,  the  difference  in  time. 

Rule.  Multiply  the  difference  of  longitude  in  degrees  and 
minutes  ly  4,  and  the  product  will  de  the  differ e7ice  of  time  in 
minutes  and  seconds,  which  may  he  reduced  to  hours. 

If  one  place  be  in  east,  and  the  other  in  west  longitude,  the  difference  of  longitude 
is  found  by  adding  them,  and  if  the  sum  be  greater  than  180^,  it  mubt  be  subtracted 
from  360^ 

Examples  for  Peactice. 
2.  New  York  is  74°  1'  and  Cincinnati  84°  24'  west  longi- 
tude ;  what  is  the  difference  of  time  ?  Ans.  41  min.  32  sec. 
jj  3.  The  Cape  of  Good  Hope  is  18°  28'  east,  and  the  Sand- 
wich Islands  155°  west  longitude ;  what  is  the  difference  of 
time?  Ans.  11  h.  33  min.  52  sec. 

4.  Washington  is  77°  1'  west,  and  St.  Petersburgh  30°  19 
east  longitude ;  what  is  their  difference  of  time  ? 

Ans.  7  h.  9  min.  20  sec. 

5.  If  Pekin  is  118°  east,  and  San  Francisco  122°  west 
longitude,  what  is  their  difference  of  time  ? 

6.  If  a  message  be  sent  by  telegraph  without  any  loss  of 
time,  at  12  M.  from  London,  0°  0'  longitude,  to  Washington, 
77°  1'  west,  what  is  the  time  of  its  receipt  at  Washington? 

Since  the  sun  appears  to  move  trom  east  to  west,  when  it  Is  exactly  12  o'clock  at 
one  place,  it  will  be  past  12  o'clock  at  all  places  east,  and  btfore  12  at  all  places  west. 
Hence,  knowing  the  difference  of  time  between  two  places,  and  the  exact  time  at 
one  of  them,  the  exact  time  at  the  other  will  be  found  by  adding  their  difference  to 
the  given  time,  if  it  be  east,  and  by  subtracting  if  it  be  toest. 

Ans.  6  h.  51  min.  56  sec.  A.  m. 
Give  explanation.    Rale. 


COMPOUKD    KUMBEES.  197 

^  7.  A  steamer  arrives  at  Halifax,  63°  36'  west,  at  4  o'clock, 
p.  M. ;  the  fact  is  telegraphed  to  St.  Louis,  90°  15'  west, 
without  loss  of  time ;  what  is  the  time  of  its  receipt  at  St. 
Louis  ?  Ans.  2  h.  13  min.  24  ^ec,  P.  m. 

<''8.  If,  at  a  presidential  election,  the  voting  begin  at  sun- 
rise and  end  at  sunset,  how  much  sooner  will  the  polls  open 
and  close  at  Eastport,  Me.,  67°  west,  than  at  Astoria,  Ore- 
gon, 124°  west  ?  Ans.  3  h.  48  min. 

9.  When  it  was  1  o'clock,  A.  m.,  on  the  first  day  of  Jan- 
uary, 1859,  at  Bangor,  Me.,  68°  47'  west,  what  was  the  time 
at  the  city  of  Mexico,  99°  5'  west  ? 

Ans.  Dec.  31,  1858,  58  min.  48  sec.  past  10,  P.  M. 

Case  IL 

226.  To  find  the  difference  of  longitude  between 
two  places,  when  the  difference  of  time  is  known. 

1.  If  the  difference  of  time  between  New  York  and  Cin- 
cinnati be  41  min.  32  sec,  what  is  the  difference  of  longitude  ? 

OPERATION.  Analysis.     Since  4  minutes  of  time 

min.        sec.  make  a  difference  of  1°  of  longitude,  and 

4  )  41          32  4  seconds  of  time,  a  difference  of  V  of 

-.^o        nor     A  longitude,  there  will  be  \  as  many  de- 

'  '          grees  of  longitude  as  there  are  minutes 

of  time,  and  \  as  many  minutes  of  longitude  as  there  are  seconds  of 

time. 

EuLE.  Reduce  the  difference  of  time  to  mimctes  and  sec- 
onds, and  then  divide  hy  4  ;  the  quotient  will  he  the  difference 
in  longitude,  in  degrees  and  minutes. 

2.  What  is  the  difference  of  longitude  between  the  Cape 
of  Good  Hope  and  the  Sandwich  Islands,  if  the  difference 
of  time  be  11  h.  33  min.  52  sec.  ?  Ans.  173°  28'. 

3.  What  is  the  difference  of  longitude  between  Washings 
ton  and  St.  Petersburg,  if  their  difference  of  time  be  7  h. 
9  min.  20  sec?  Ans.  107°  20'. 

Case  II  is  what  ?    Give  explanation.    Rule. 


198  DUODECIMALS. 

4.  When  it  is  half  past  4,  p.  m.,  at  St.  Petersburg,  30°  19' 
east,  it  is  32  min.  36  sec.  past  8,  a.m.,  at  New  Orleans, 
west;  what  is  the  difference  of  longitude?  Ans.  119°  21', 

5.  The  longitude  of  New  York  is  74°  1'  west.  A  sea  cap- 
tain leaving  that  port  for  Canton,  with  New  York  time, 
finds  that  his  chronometer  constantly  loses  time.  What  is 
his  longitude  when  it  has  lost  4  hours  ?  8  h,  40  min.?  13  h. 
25  min.  ?      Ans.  14°  1'  west ;  55°  59'  east ;  127°  14'  east. 

6.  AVhen  the  days  are  of  equal  length,  and  it  is  noon  on 
the  1st  meridian,  on  what  meridian  is  it  then  sunrise  ?  sun- 
set? midnight?  ^725.  90°  west;  90°  east;  180°  east  or  west. 

DUODECIMALS. 

227.  Duodecimals  are  the  divisions  and  subdivisions  of 
a  unit,  resulting  from  continually  dividing  by  12,  as*  -j^,  y^, 
Th'^,  etc.  In  practice,  duodecimals  are  applied  to  the  meas- 
urement of  extension,  the  foot  being  taken  as  the  unit. 

If  the  foot  be  divided  into  12  equal  parts,  the  parts  are 
called  inches,  or  primes ;  the  inches  divided  by  12  give  sec- 
onds ;  the  seconds  divided  by  12  give  thirds ;  the  thirds 
divided  by  12  give  fourths ;  and  so  on. 

From  these  divisions  of  a  foot  it  follows  that 

V    (Inch  or  Prime) is    -^    of  a  foot. 

1"  (Second)  or  ^  of  ^ "  ^ir  of  a  foot 

1'"  (Third)  or  ^^  ot  -^  ot  ^. .  "  ^^  o^  a  foot. 

Table. 

12  Fourtlis,  marked  (""),  make  1  Third marked  1'" 

12  Thirds  "     1  Second  "       1" 

12  Seconds  **      1  Prime,  or  Inch,      "       1' 

13  Primes,  or  Inches,  "     1  Foot  " '     ft. 
Scale— uniformly  12. 

The  marks  ',  ",  '",  "",  are  called  indices. 

What  arc  duodecimals?  To  what  applied?  Explain  the  divisions 
of.  the  foot.     Repeat  the  table. 


COMPOUND     I?^UMBERS.  '  199 

Duodecimals  are  really  common  fractions,  and  can  always  be  treated  as  such ;  bat 
usually  their  denominators  are  not  expressed,  and  they  are  treated  as  compoand 
numbers. 

Addition  and  Subtraction  of  Duodecimals 

228,  We  add  and  subtract  duodecimals  the  same  as  other 
compound  numbers. 

Examples  for  Practice. 

1.  Add  13  ft.  4'  8",  10  ft.  6'  7",  145  ft.  9'  11". 

Ans.  169  ft.  9'  2". 

2.  Add  179  ft.  11'  4",  245  ft.  1'  4",  3  ft.  9'  9". 

Ans,  428  ft  10'  5". 

3.  From  25  ft.  6'  3"  take  14  ft  9'  8".    A71S.  10  ft  8'  7". 

4.  From  a  board  15  ft  7'  6"  in  length,  3  ft.  8'  11"  were 
sawed  off;  what  was  the  length  of  the  piece  left? 

Ans.  11  ft  10'  7". 

Multiplication  of  Duodecimals. 

229.  Length  multiplied  by  breadth  gives  surface,  and 
surface  multiplied  by  thickness  gives  sohd  contents  ( 198 ). 

1.  How  many  square  feet  in  a  board  11  feet  8  inches  long 
and  2  feet  7  inches  wide  ? 

Analysis.    We  first  multiply  hj  the  7'. 

7  twelfths  times  8  twelfths  equals  56  one 

hundred    forty-fourths,   which    equals  4 

twelfths  and  8  one  hundred  forty-fourths. 

We  write  the  8  144ths  —  marked  with  two 

indices — to  the  right,  and  add  the  4  12th6 

30  ft       1'     8"  ^  *^®  ri^^t  product.    7'  times  11  equals 

77',  which  added  to  4'  equals  81',  equal  to 

6  feet  and  9'.    We  write  the  9'  under  the 

inches,  or ;  12ths,  and  the  6  under  the   feet,  or  units.     2  times  8 

equals  16',  or  1   foot  and  4'.    We  write  the  4'  under  the  9',  and 

add  the  1  foot  to  the  next  product.    2  times  11  feet  are  22  feet,  and 

1  foot  added  make  23  feet,  which  we  write  under  the  6  feet.    Adding 

How  are  duodecimals  added  and  suhtracted  ?  Give  analysis  of  ex- 
ample 1. 


OPKRATION 

lift 

8' 

2 

7' 

6  ft 

9' 

8" 

23 

4' 

k 


200  DUODECIMALS. 

these  partial  products,  and  we  have  30  ft.  V  and  8"  for  the  entire 
product. 

It  will  be  seen  from  the  above  that  the  number  of  indiees  to  every 
product  of  any  two  factors  is  equal  to  the  sum  of  the  indices  of  those 
factors  ;  thus :  7'  x  8'=56"  ;  4"  x  5"'=20""'. 

EuLE.  I.  Wrile  the  several  terms  of  the  multiplier  urder 
the  corresponding  terms  of  the  multiplicand. 

II.  Multiply  each  term  of  the  multiplicand  hy  each  term  of 
the  multiplier f  heginning  with  the  lowest  term  in  each,  and 
call  the  product  of  any  two  denominations  the  denomination 
denoted  hy  the  sum  of  their  indices,  carrying  1  for  every  12. 

III.  Add  the  partial  products,  carrying  1  for  every  12 ; 
their  sum  will  he  the  required  answer. 

Examples  for  Peactice. 

2.  How  many  square  feet  in  a  board  13  ft.  9'  long  and  11 
wide  ?  Ans.  12  ft.  7'  3". 

3.  How  many  square  feet  in  a  stock  of  4  boards,  each 
11  ft.  9'  long  and  1  ft.  3'  wide  ?  A7is.  58  ft.  9'. 

4.  How  many  square  yards  of  plastering  on  the  walls  of  a 
room  12  ft.  11'  square,  and  9  ft.  3'  bigb,  allowing  for  two 
windows  and  one  door,  each  6  ft.  2'  high  and  2  ft.  4'  wide  ? 

Ans,  48  sq.  yd.  2  ft.  9'. 

5.  How  many  solid  feet  in  a  mow  of  hay  30  ft.  4'  long, 
25  ft.  6'  wide,  and  12  ft  5'  high  ?        Ans.  9604  ft.  3'  6". 

6.  How  many  cords  in  a  pile  of  wood  18  ft.  6'  long,  12  ft. 
wide,  and  5  ft.  6'  high  ?  Ans.  ^  cords  69  ft. 

7.  How  many  cubic  yards  of  earth  must  be  removed  in 
digging  a  cellar  36  ft.  10'  long,  22  ft.  3'  wide,  and  5  ft.  2' 
deep  ?  Ans.  156  cu.  yd.  22  ft.  3'  7". 

8.  What  would  it  cost  to  plaster  a  wall  32  ft.  8'  long  and 
9  ft.  high,  at  17  cents  per  square  yard?  Ans,  $5.55 J. 

9.  How  many  yards  of  carpeting,  27'  wide,  will  be  re- 
quired to  cover  a  floor  48  ft.  long  and  33  ft.  9'  wide  ? 

Ans.  240  yards. 

Qive  the  rale. 


COMPOUND     NUMBEES.  201 

Divisioiq-  OF  Duodecimals. 

230.  1.  A  flagstone,  3  ft.  9'  wide,  has  a  surface  of  20  ft 
11' 3";  what  is  its  length? 

OPERATioif.  Analysis.    We  divide 

3  ft.  9'  )  20  ft  11'     3"  (  5  ft  T.        the  surface  by  the  width 

"l^g  9'  to  obtain  the  length.    The 

—  divisor  is  something  more 

^  ^3  than  3  ft.,  and  to  obtain 

2  2      3  the  first  quotient  figure,  wo 

consider  how  many  times 
3  ft.  and  something  more  is  contained  in  nearly  21  ft.  (20  ft.  11') ;  we 
estimate  it  to  be  5  times,  and  multiplying  the  divisor  by  this  quotient 
figure,  we  have  18  ft.  9',  which,  subtracted  from  20  ft.  11',  leaves 
2  ft.  2',  to  which  we  bring  down  3",  the  last  term  of  the  dividend.  We 
next  seek  how  many  times  the  divisor  is  contained  in  this  remainder, 
and  find  by  trial  the  quotient  7' ;  multiplying  the  divisor  by  this  figure, 
we  obtain  2  ft.  2'  3",  and  there  is  no  remainder. 

KuLE.  I.  Write  the  divisor  on  the  left  hand  of  the  dividend, 
as  in  simple  numbers, 

II.  Find  the  first  term  of  the  quotient  either  hy  dividing 
the  first  term  of  the  dividend  by  the  first  term  of  the  divisor, 
or  hy  dividing  the  first  two  terms  of  the  dividend  by  the  first 
two  terms  of  tlie  divisor;  m,ultiply  the  divisor  by  this  term 
of  the  qtwtienty  subtract  the  product  from  the  corresponding 
terms  of  the  dividend,  and  to  the  remainder  bring  doicn  an- 
other term  of  the  divisor. 

III.  Proceed  in  like  manner  till  there  is  no  remainder,  or 
till  a  quotient  has  been  obtained  sufficiently  exact. 

Examples  for  Practice. 

2.  Divide  44  ft  5'  4"  by  16  ft.  8'.  Ans,  2  ft  8'. 

3.  The  squai'e  contents  of  a  walk  are  184  ft  3'  and  tlje 
length  is  40  ft  11'  4"  ;  what  is  the  width  ?    Ans.  4  ft  6'. 

4.  A  blanket  whose  square  contents  are  14  ft.  6',  is  to  be 
lined  with  cloth  2  ft  7'  wide  ;  how  much  in  length  will  be 
required  ? 

Give  analysis  of  example  1.     Rule. 
9* 


202  PKOMISCUOUS     EXAMPLES. 

5.  A  block  01  granite  contains  64  ft.  2'  5" ;  its  width  is 
2  ft.  6',  and  its  thickness  3  ft.  7' ;  what  is  its  length  ? 

Am,  7  ft.  2'. 

Since  the  solid  contents  are  the  product  of  the  three  dimensions,  we  divide  the 
solid  contents  by  any  two  denominations  or  by  their  product,  to  obtain  the  other 
dimension. 

Peomiscuous  Examples. 

1.  In  115200  grains  Troy,  how  many  pounds  ? 

2.  In  365  da.  6  h.  48  min.  46  sec,  how  many  seconds  ? 

Ans.  31556926. 

3.  A  man  wishes  to  ship  1560  bushels  of  potatoes  in  bar- 
rels containing  3  bu.  1  pk.  each  ;  how  many  barrels  will  be 
required?  Ans,  480. 

4.  Eeduce  295218  inches  to  miles. 

5.  Reduce  456575  grains  to  pounds,  apothecaries'  weight. 

Ans.  79ib3i  I3  l3l5gr. 

6.  How  many  sheets  in  3  reams  of  paper  ? 

7.  What  is  the  value  of  4  piles  of  wood,  each  20  ft.  long, 
6  ft.  wide,  and  10  ft.  high,  at  $3.25  per  cord  ?    Ans.  $121.87|. 

8.  How  many  bottles,  each  holding  1  qt.  1  gi.,  can  be  filled 
from  a  barrel  of  cider?  A71S,  112. 

9.  At  $26. 40  per  sq.  rd.  for  land,  what  will  be  the  cost  of  a 
village  lot  SJ  rd.  long,  and  ^  rd.  wide?       Ans.  $980.10. 

10.  Divide  259  A.  50  P.  of  land  into  36  equal  lots. 

Ans.  7  A.  32^  P. 

11.  How  many  times  can  a  box  holding  4  bu.  3  pk.  2  qt. 
be  filled  from  336  bu.  3  pk.  4  qt.  ?  Ans.  70. 

12.  What  is  the  value  of  .875  of  a  gallon  ? 

13.  What  part  of  a  mile  is  116  rd.  2  yd.  ?  Ans.  •^. 

14.  What  part  of  2  days  is  13  h.  26  min.  24  sec.  ? 

'  15.  From  26  A.  80  P.  of  land,  5  A.  120  P.  were  sold  ;  what 
part  of  the  whole  piece  remained  unsold  ?  Am.  ■^. 

16.  What  is  the  difference  between  -J  of  a  pound  sterling 
and  5|-  pence  ?  ^«*.  lis.  6id. 

17.  What  is  the  sum  of  ^  of  a  yard,  ^^  of  a  foot,  and  \  of 
an  inch  ?  An^.  7  inches. 


PROMISCUOUS     EXAMPLES.  203 

18.  Reduce  3  cwt.  1  qr.  7  lb.  of  coal  to  the  decimal  of  a 
long  ton.  Ans.  .165625. 

19.  Benjamin  Franklin  was  born  Jan.  18,  1706,  and 
George  Washington  Feb.  22,  1732  ;  how  much  older  was 
Franklin  than  Washington  ?  Ans.  26  jv.  1  mo.  4  da. 

20.  The  longitude  of  Boston  is  71°  4'  west,  and  that  of 
Chicago  87°  30'  west ;  when  it  is  12  m.  at  Boston,  what  is  the 
time  in  Chicago  ?  Ans.  10  h.  54  min.  16  sec.  A.  m. 

21.  If  the  difference  of  time  between  New  York  and  New 
Orleans  be  1  h.  4  sec,  what  is  the  difference  in  longitude  ? 

Ans.  15°  1'. 

22.  Add  f  of  a  mile,  ^  of  a  mile,  and  -f^  of  a  rod  to- 
gether. Ans.  233  rd.  8  ft.  3  in. 

23.  If  a  bushel  of  barley  cost  $.80,  what  will  20  bu.  3  pk. 
6qt.  cost?  Ans.  $16.75. 

24.  What  is  the  value  of  .875  of  a  gross  ?    Ans.  10|-  doz. 

25.  How  many  acres  in  a  field  56^  rods  long,  and  24.6 
rods  wide  ?  Ans.  8  A.  109.9  P. 

26.  How  many  perches  of  masonry  in  the  wall  of  a  cellar 
which  is  20  feet  square  on  the  inside,  8  feet  high,  and  IJ  feet 
in  thickness  ?  Ans.  44.6  +  . 

27.  A,  B,  and  C  rent  a  farm,  and  agree  to  work  it  upon 
shares  ;  they  raise  640  bu.  3  pk.  of  grain,  which  they  divide 
as  follows  :  one-fourth  is  given  for  the  rent ;  of  the  remain- 
der A  takes  10^  bu.  more  than  one- third,  after  which  B  takes 
one-half  of  the  remainder  less  7  bushels,  and  C  has  what  is 
left ;  how  much  is  C's  share  ?      Ans.  161  bu.  3  pk.  6  qt. 

28.  What  is  the  value  in  Troy  weight  of  13  lb.  8  oz.  Avoir- 
dupois weight  ?  Ans.  16  lb.  4  oz.  17  pwt.  12  grt 

29.  If  154  bu.  1  pk.  6  qt.  cost  $173.74,  how  much  will  1.5 
bushels  cost  ?  Ans.  $1,687. 

30.  What  is  the  value  of  .0125  of  a  ton  ?      Ans.  25  lbs. 

31.  AYhat  fraction  of  3  bushels  is  -^\  of  2  bu.  3  pk.  ? 

Ans.  -^. 

32.  How  many  wine  gallons  in  a  water  tank  4  feet  long, 
3^  feet  wide,  and  1  ft.  8  in.  deep  ;  Ans.  174j^. 


204  PROMISCUOUS    EXAMPLES. 

33.  How  many  bushels  will  a  bin  contain  that  is  7|  feet 
square,  and  6  ft.  8  in.  deep?  Ans.  301.339 4-bu. 

34.  How  much  must  be  paid  for  lathing  and  plastering 
overhead  a  room  36  feet  long  and  20  feet  wide,  at  26  cents 
a  square  yard  ? 

35.  How  many  shingles  will  it  take  to  cover  the  roof  of  a 
building  46  feet  long,  each  of  the  two  sides  of  the  roof  being 
20  feet  wide,  allowing  each  shingle  to  be  4  inches  wide,  and 
to  lie  5  inches  to  the  weather  ?  Ans.  13248. 

36.  John  Young  was  born  at  a  quarter  before  4  o'clock,  A. 
M.,  Sept.  4, 1836  ;  what  will  be  his  age  at  half -past  6  o'clock, 
p.  M.,  April  20,  1864  ?     Ans.  27  yr.  7  mo.  16  da.  14  h.  45  min. 

37.  How  many  cubic  yards  of  earth  were  removed  in  dig- 
ging a  cellar  28  ft.  9'  long,  22  ft.  8'  wide,  and  7  ft.  6'  deep  ? 

A71S.  181-gJj  cu.  yd. 

38.  What  will  30  bu.  54  lb.  of  wheat  cost,  at  $1.3  7 J  per 
bushel?  Ans.  $42.4875. 

39.  How  many  square  yards  of  cai-peting  will  it  take  to 
cover  a  floor  24  ft.  8'  long  and  18  ft.  6'  wide  ?     Ans.  60^. 

40.  What  is  the  cost  of  54  bu.  8  lb.  of  barley,  at  84  cents 
per  bushel?  Ans.  $45.60. 

41.  What  is  the  depth  of  a  lot  that  has  120  feet  front,  and 
contains  18720  square  feet  ? 

42.  How  many  steps  of  30  inches  each  must  a  person 
take  in  walking  21  miles  ? 

43.  How  long  will  it  require  one  of  the  heavenly  bodies  to 
move  through  a  quadrant,  if  it  move  at  the  rate  of  3'  12" 
per  minute  ?  Ans.  1  da.  4  h.  7  min.  30  sec. 

44.  How  many  times  will  a  wheel,  9  ft.  2  in.  in  circum- 
ference, turn  round  in  going  65  miles  ? 

45.  If  a  man  buy  10  bushels  of  chestnuts,  at  $5.00  per 
bushel,  dry  measure,  and  sell  the  same  at  22  cents  per  quart, 
liquid  measure,  how  much  is  his  gain?  A7is.  $31.92. 

46.  What  will  it  cost  to  build  a  wall  240  feet  long,  6  feet 
high,  and  3  feet  thick,  at  $3.25  per  1000  bricks,  each  brick 
being  8  inches  long,  4  inches  wide,  and  2  inches  thick  ? 

Ans.  $379.08. 


PERCEKTAGE. 


205 


PERCENTAGE. 

231.  Per  cent,  is  a  term  derived  from  the  Latin  words 
per  centum,  and  signifies  by  the  hundred,  or  hundredths, 
that  is,  a  certain  number  of  parts  of  each  one  hundred  parts, 
of  whatever  denomination.  Thus,  by  5  per  cent,  is  meant  5 
cents  of  every  100  cents,  15  of  every  $100,  5  bushels  of  every 
100  bushels,  etc.  Therefore,  5  per  cent,  equals  5  hundredths 
— .05=:yfo=r-^.    8  per  cent,  equals  8  hundredths =.08= 

232.  Percentag-e  is  such  a  part  of  a  number  as  is  indi- 
cated by  the  per  cent. 

233.  The  Base  of  percentage  is  the  number  on  which 
the  percentage  is  computed. 

234.  Since  per  cent,  is  any  number  of  hundredths,  it  is 

usually  expressed  in  the  form  of  a  decimal ;  but  it  may  be 

expressed  either  as  a  decimal  or  a  common  fraction,  as  in 

the  following 

Table. 
Decimals.       Common  Fractions.    Lowest  Terms. 


1    per  cent. 

= 

.01 

= 

T^iy 

= 

TcHT 

2    per  cent. 

= 

.02 

= 

TW^ 

= 

^ 

4    per  cent. 

= 

.04 

= 

T¥ir 

= 

1^ 

5    per  cent. 

= 

.05 

= 

TlTff 

= 

1^ 

6    per  cent. 

= 

.06 

= 

1%-S 

= 

t 

7    per  cent. 

= 

.07 

= 

10  0 

= 

T^ 

8    per  cent. 

= 

.08 

= 

TTTD" 

r= 

^ 

10    per  cent. 

= 

.10 

= 

uny 

= 

^ 

16    per  cent. 

= 

.16 

= 

^% 

= 

A 

20    percent. 

= 

.20 

= 

"J? 

= 

i 

25    per  cent. 

= 

.25 

= 

1^ 

= 

i 

50    per  cent. 

= 

.50 

= 

tVo 

= 

i 

100    per  cent. 

= 

1.00 

= 

m 

= 

1 

125    per  cent. 

= 

1.25 

= 

m 

= 

f 

\  per  cent. 

= 

.005 

r= 

unnr 

= 

^(T 

1  per  cent. 

= 

.0075 

= 

Tff'imj 

= 

■^ 

12|^  per  cent. 

= 

.125 

= 

^¥^ 

= 

i 

16^  per  cent. 

= 

.1625 

= 

tWA 

= 

M 

What  is  meant  by  per  cent.?  From  what  is  the  term  derived? 
What  is  percentage?  What  is  the  base  of  percentage?  How  is  per 
cent,  expressed  ? 


206  percentage. 

Examples  for  Practice. 

1.  Express  decimally  3  per  cent. ;  6  per  cent. ;  9  per  cent. ; 
14  per  cent. ;  24  per  cent. ;  40  per  cent. ;  122  J  per  cent. ;  150 
per  cent. 

2.  Express  decimally  6J  per  cent.;  8f  per  cent.;  33^  per 
cent.;  7i  per  cent.;  lOf  per  cent.;  9|  per  cent;  103 J  per 
cent. ;  225  per  cent. 

3.  Express  decimally  J  per  cent.;  f  per  cent.;  f  per  cent.; 
f  percent.;  |  per  cent;  IJ  per  cent;  2^  per  cent;  4^  per 
cent;  5}  per  cent;  7^  per  cent;  12^  per  cent;  25|  per 
cent. 

4.  Express  in  the  form  of  common  fractions,  in  their  low- 
est terms,  6  per  cent;  8  per  cent;  12  percent;  14|^  per 
cent;  18-|  per  cent.;  21f  per  cent;  31  J-  per  cent ;  37^  per 
cent.;  40f  per  cent;  112  per  cent;  225  per  cent 

Case  I. 
235.  To  find  the  percentage  of  any  number. 

1.  A  man,  having  $125,  lost  4  per  cent  of  it ;  how  many 
dollars  did  he  lose  ? 

OPERATION. 

$125  AiTALTSis.    Since  4  per  cent,  is  yj^  =  .04,  he  lost 

Q4  .04  of  $125,  or  $125  x  .04  =  $5.     Or,  4  per  cent,  is 

•^^  T^ir  =  -iz^  and  ^V  of  $125  ^  $5. 

Rule.  Multiply  the  given  number  or  quantity  by  the  rate 
per  cent,  expressed  decimally,  and  point  off  as  in  decimals.   Or, 

Take  such  a  part  of  the  given  number  as  tJie  number  express- 
ing the  rate  is  part  of  100. 

Examples  for  Practice. 

2.  What  is  6  per  cent  of  $320  ?  Ans.  $19.20. 

3.  What  is  8  per  cent  of  $327.25  ?  Ans.  $26.18. 

Case  I  is  what?    Give  explanation.    Rule. 


PEECEJfTAGE.  207 

4.  What  is  7}  per  cent,  of  $56.75  ?  Ans.  $4.11^. 

5.  What  is  12J  per  cent,  of  2450  pounds  ? 

Ans,  306.25  pounds. 

6.  What  is  6|  per  cent,  of  19072  bushels? 

Aiis.  1287.36  bushels. 

7.  What  is  33^  per  cent  of  846  gallons  ? 

Ans.  282  gallons. 

8.  What  is  9|  per  cent,  of  275  miles?    Ans,  26.95  miles. 

9.  What  is  14  per  cent,  of  450  sheep  ? 

10.  What  is  50  per  cent,  of  1240  men? 

11.  What  is  105  per  cent,  of  $5760  ?  Ans,  $6048. 

12.  What  is  175  per  cent,  of  S12967? 

13.  What  is  25  per  cent,  of  |  ? 

25  per 'cent,  equals  ^^^  =  \,  and  lx^  =  -^,  Ans. 

14.  What  is  15  per  cent,  of  f  ?  Ans.  y^. 

15.  What  is  2^  per  cent,  of  6f  ?  Ans.  i. 

16.  What  is  33-J-  per  cent,  of  -3^  ?  Ans.  -^. 

17.  What  is  84  per  cent,  of  7i?  Ans.  6-^^. 

18.  Find  |  per  cent,  of  $40.80.  Ans.  $.306. 

19.  Find  If  per  cent,  of  $15.60.  Ans.  $.26. 

20.  A  farmer,  having  760  sheep,  kept  25  per  cent,  of 
them,  and  sold  the  remainder ;  how  many  did  he  sell  ? 

21.  A  man  has  a  capital  of  $24500;  he  invests  18  per 
cent,  of  it  in  bank  stock,  30  per  cent,  of  it  in  railroad  stocks, 
and  the  remainder  in  bonds  and  mortgages  ;  how  much  does 
he  invest  in  bonds  and  mortgages  ?  Ans.  $12740. 

22.  A  speculator  bought  1576  barrels  of  apples,  and  upon 
opening  them  he  found  12^  per  cent,  of  them  spoiled  ;  how 
many  barrels  did  he  lose  ? 

23.  Two  men  engaged  in  trade,  each  with  $2760.  One 
of  them  gained  33-J-  per  cent,  of  his  capital,  and  the  other 
gained  75  per  cent. ;  how  much  more  did  the  one  gain  than 
the  other?  ^ws.  $1150. 

24.  A  man,  owning  f  of  an  iron  foundry,  sold  35  per  cent, 
of  his  share ;  what  part  of  the  whole  did  he  sell,  and  what 
part  did  he  still  own  ?  Ans.  He  still  owned  ^|. 


208  PERCENTAGE. 

25.  A  owed  B  $575.40;  he  paid  at  one  time  40  per  cent, 
of  the  debt ;  afterward  he  paid  25  per  cent,  of  the  remain- 
der; and  at  another  time  12^  per  cent,  of  what  he  owed 
after  the  second  payment ;  how  much  of  the  debt  did  he 
stiU  owe  ?  Ans,  $226.56|. 

Case  II. 

236.  To  find  what  per  cent,  one  number  is  of 
another. 

1.  A  man,  having  $125,  lost  $5;  what  per  cent,  of  his 
money  did  he  lose  ? 

OPERATION.  ^  Analysis.    Multiply  the 

5_i_125=.04=:4  per  cent.  base  by  the  rate  per  cent. 

Or,  to  obtain  the   percentage 

yf^=-g«^=.04=4  per  cent:  (235)  ;  conversely,  divide 

the  percentage  by  the  base 
to  obtain  the  rate  per  cent.  Or,  since  $125  is  100  per  cent,  of  his 
money,  $5  is  yf^,  equal  to  -^^  of  100  per  cent.,  which  is  4  per  cent. 

EuLE.  Divide  the  percentage  ly  the  base,  and  the  quotient 
will  be  the  rate  per  cent,  expressed  decimally.    Or, 

Take  such  a  part  of  100  as  the  percentage  is  part  of  the 
base. 

Examples  for  Practice, 

2.  What  per  cent,  of  $450  is  $90  ?  Ans.  20. 

3.  What  per  cent,  of  $1400  is  $175  ?  Ans,  12i. 

4.  What  percent,  of  $750  is  $165? 

5.  What  per  cent,  of  $240  is  $13.20  ?  Ans.  5f 

6.  What  per  cent,  of  $2  is  15  cents  ? 

%■  What  per  cent,  of  G  bushels  1  peck  is  4  bushels  2  pecks 
6  quarts?  '  Ans.  75  per  cent 

8.  What  per  cent,  of  15  pounds  is  5  pounds  10  ounces 
avoirdupois  weight  ?  -4ws.  37-J-  per  cent. 

9.  What  per  cent,  of  250  head  of  cattle  is  40  head  ? 

Case  II  is  what  ?    G  ive  explanation.     Rule. 


PERCENTAGE.  209 

10.  From  a  hogshead  of  sugar  containing  760  pounds,  100 
pounds  were  sold  at  one  time,  and  90  pounds  at  another ; 
what  per  cent,  of  the  whole  was  sold  ? 

11.  A  man,  having  600  acres  of  land,  sold  J  of  it  at  one 
time,  and  -J  of  the  remainder  at  another  time ;  what  per 
cent,  remained  unsold  ?  Ans,  50  per  cent. 

Case  III. 

237.  To  find  a  nnmber  when  a  certain  per  cent, 
of  it  is  given. 

1.  A  man  lost  $5,  which  was  4  per  cent,  of  aU  the  money 
he  had ;  how  much  had  he  at  first  ? 

OPEKATiON.  Analysis.    We  are  here  required  to 

^5_i_.04=$125.  filial  the  base,  of  wMcli  $5  is  the  per- 

Qp  centage.    Now,  percentage  equals  base 

4^  V  1 00 ^1 25  multiplied  by  the  rate  per  cent. ;  con- 

*  *  Tersely,  base  equals  percentage  divided 

by  rate  per  cent.  Or,  $5  Ls  4  per  cent,  of  all  he  had  ;  |^  of  $5,  or  f , 
equals  1  per  cent,  of  all  he  had,  and  100  times  f  equals  100  per  cent., 
or  all  he  had. 

Rule.  Divide  the  percentage  ty  the  rate  per  cent.,  ex- 
pressed decimally,  and  the  quotient  will  be  the  base,  or  num- 
ber required.     Or, 

Talce  as  many  times  100  as  the  percentage  is  times  the  rate 

per  cent. 

Examples  for  Practice. 

2.  16  is  8  per  cent,  of  what  number  ?  Ans.  200. 

3.  42  is  7  per  cent,  of  what  number  ? 

4.  75  is  12^  per  cent,  of  what  number?  Ans.  600. 

5.  33  is  2|  per  cent,  of  what  number  ?  Ans.  1200. 

6.  $281.25  is  37i  per  cent,  of  what  sum  of  money? 

Ans.  $756. 

7.  A  farmer  sold  50  sheep,  which  was  20  per  cent,  of  his 
whole  flock  ;  how  many  sheep  had  he  at  first  ? 

Case  III  is  what  ?    Give  explanation.    Rule. 


210  PERCEl^TAGE. 

8.  I  loaned  a  man  a  certain  sum  of  money ;  at  one  time 
he  paid  me  $59.75,  which  was  12J  per  cent,  of  the  whole 
sum  loaned  to  him ;  how  much  did  I  loan  him  ? 

9.  A  merchant  invested  $975  in  dry  goods,  which  was  15 
per  cent,  of  his  entire  capital ;  what  was  the  amount  of  his 
capital  ?  Ans.  $6500. 

10.  If  a  man,  owning  40  per  cent,  of  an  iron  foundry, 
sell  25  per  cent,  of  his  share  for  $1246.50,  what  is  the  value 
of  the  whole  foundry  ?  Ans.  $12465. 

11.  A  produce  buyer,  having  a  quantity  of  corn,  bought 
2000  bushels  more,  and  he  found  that  this  purchase  was  40 
per  cent,  of  his  whole  stock ;  how  much  had  he  before  he 
bought  this  last  lot  ?  Ans.  3000  bushels. 

Case  IV. 

208.  To  find  a  number  when  the  number,  in- 
creased by  a  certain  per  cent,  of  itself,  is  given. 

1.  A  man's  income  this  year  is  $525,  which  is  5  per  cent, 
inore  than  it  was  last  year ;  what  was  it  last  year  ? 

OPERATION.  Analysis.    Since  his  income 

$525  -j-  1.05  =  $500.  this  year  is  .05  more  than  it 

was  hist  year,  this  year's  in- 
come must  be  1.05  times  the  income  of  last  year ;  therefore  divide  this 
year's  income  by  1.05  and  it  gives  the  income  of  last  year. 

KuLE.  Divide  the  amount  ly  1  plus  the  rate  expressed 
decimally,  and  the  quotient  will  be  the  base  or  number  re- 
quired.   Or, 

Take  as  many  times  100  as  the  amount  is  times  1  plus  the 
rate  per  cent. 

What  is  Case  IV  ?    Explanation  ?    Rule. 


peecentage.  211 

Examples  for  Practice. 

2.  What  number  increased  by  18  per  cent,  of  itseK  is 
equal  to  1475  ?  "  Ans.  1250. 

3.  A  merchant  sells  broadcloth  for  $4  per  yard,  and  there- 
by makes  25  per  cent. ;  what  did  the  cloth  cost  him  ? 

Ans.  $3.20. 

4.  A  expended  a  certain  sum  for  a  house,  and  15  per  cent. 
of  the  purchase  price  on  repairs,  and  then  found  that  the 
whole  cost  was  $6900.     What  was  the  purchase  price  ? 

5.  A  certain  manufacturing  company  have  sold  1432,250 
worth  of  goods,  which  is  8^  per  cent,  more  than  they  sold 
last  year.    How  much  did  they  sell  last  year  ?  Ans.  1400,000. 

6.  A  merchant  bought  a  stock  of  goods,  and  paid  4^  per 
cent,  of  the  purchase  price  for  freight,  when  he  found  that 
the  goods  cost  him  $8757.     What  was  the  purchase  price  ? 

7.  A  merchant  increased  his  capital  by  20  per  cent,  each 
year  for  two  years,  when  he  found  he  had  $9360  invested. 
How  much  had  he  at  first  ?  Ans,  $6500. 

Case  Y. 
239.   To  find  a  number  when  the  number,  dimin- 
ished by  a  certain  per  cent,  of  itself,  is  given. 

1.  A  man  lost  8  per  cent,  of  his  sheep  and  had  368  left ; 
how  many  had  he  at  first  ? 

OPERATION.  •        Analysts.    Since  the  man  lost 

368  -7-  .92  =  400.  8  per  cent,  of  his  sheep,  he  has 

92  per  cent,  left ;  hence  368  is  .92 
times  his  original  flock.  We  therefore  divide  368  by  .93  and  obtain 
the  required  number  of  sheep. 

Rule.  Divide  the  given  mimber  hy  1  minus  the  rate  ex- 
pressed decimally,  and  the  quotient  will  he  the  base  or  number 
required.     Or, 

Take  as  many  times  100  as  the  given  numler  is  times  1 
miyius  the  rate. 

Wliat  is  Case  V  ?    Explanation  ?    Rule  ? 


^13  rERCENTAGE. 

Examples  for  Practice. 

2.  What  number  diminislied  by  15  per  cent,  of  itself  is 
equal  to  340  ?  Ans.  400. 

3.  A  having  a  certain  sum  on  deposit  drew  out  20  per 
cent.,  when  he  found  he  had  $1000  left.  How  much  had  he 
on  deposit  at  first  ?  Ans.  $1250. 

4.  My  income  this  year  is  $4028,  which  is  24  per  cent, 
less  than  it  was  last  year.    How  much  was  it  last  year  ? 

5.  What  number  diminished  by  i  per  cent,  of  itself  is 
equal  to  298^^  ?  Ans.  300. 

6.  A  sells  his  horse  for  $198,  which  is  10  per  cent,  less 
than  his  asking  price,  and  his  asking  price  was  10  per  cent, 
more  than  he  cost  him  ;  what  did  the  horse  cost  him  ? 

Ans.  $200. 


COMMISSION    AND    BROKEEAGE. 

240.  An  Agent,  Factor,  or  Broker  is  a  person  wlio 
transacts  business  for  another,  or  buys  and  seUs  money, 
stocks,  notes,  etc. 

241.  Commission  is  the  percentage,  or  compensation 
allowed  an  agent,  factor,  or  commission  merchant,  for  buy- 
ing and  selling  goods  or  produce,  collecting  money,  and 
transacting  other  business. 

242.  Brokerage  is  the  fee,  or  allowance  paid  to  a  broker 
or  dealer  in  money,  stocks,  or  bills  of  exchange,  for  making 
exchanges  of  money,  buying  and  selling  stocks,  negotiating 
bills  of  exchange,  or  transacting  other  like  business. 

The  rates  of  commiesion  and  brokerage  are  not  regulated  by  law,  but  are  nsnally 
reckon^  at  a  certain  per  cent  upon  the  money  employed  in  the  transaction. 

Define  an  agent,  factor,  or  broker.  What  is  meant  by  commisBion  1 
Brokerage  ? 


COMMISSION     AND     BROKERAGE.  213 

Case  I. 

243.  To  find  the  commission  or  brokerage  on  any 
Slim  of  money. 

1.  A  commission  merchant  sells  butter  and  cheese  to  the 
amount  of  $1540  ;  what  is  his  commission  at  5  per  cent.  ? 

OPERATION.  Analysts.    Since 

$1540  X  .05  =  $77,  Ans,  the  commission  on 

Or,  T^  =  ^,  and  ^  X  $1540  =  177.     ^^  ^«  ^  ^^^«  ^^  -^^ 

it  is  $1540  X  .05  =  $77.     Or,  since  5  per  cent,  is  j^^  =  -^^  of  the  sum 
received,  the  commission  is  -^  of  $1640  =  $77. 

Rule.  Multiply  the  given  sum  ly  the  rate  per  cent,  ex- 
pressed decimally,  and  the  result  tvill  l?e  the  com?nission  or 
hrolcerage.     Or, 

Tahe  such  a  part  of  the  giveii  sum  as  the  number  express- 
ing the  per  cent,  is  part  of  100. 

Examples  for  Practice. 

2.  A  commission  merchant  sells  goods  to  the  amount  of 
$G756  ;  what  is  his  commission  at  2  per  cent.  ? 

Ans.  $135.12. 

3.  What  commission  must  be  paid  for  collecting  $17380, 
at  3^  per  cent.  ?  Ans,  $608.30. 

4.  An  agent  in  Chicago  purchased  4700  bushels  of  wheat 
at  75  cents  a  bushel ;  what  was  his  commission  at  \\  per 
cent,  on  the  purchase  money  ? 

5.  A  broker  in  New  York  exchanged  $25875  on  the  Suf- 
folk Bank,  Boston,  at  \  per  cent. ;  how  much  brokerage  did 
he  receive  ?  Ans.  $64.6875. 

6.  An  auctioneer  sold  at  auction  a  house  for  $3284,  and 
the  furniture  for  $2176.50  ;  what  did  his  fees  amount  to  at 
2:J-  per  cent.  ? 

7.  A  broker  negotiates  a  bill  of  exchange  of  $2890  for 
\  per  cent,  commission ;  how  much  is  his  brokerage  ? 

Ans.  $23.12. 
Case  I  is  what  ?     Give  explanation.    Rule. 


214  PERCENTAGE. 

8.  An  agent  buys  for  a  manufacturing  company  26750 
pounds  of  wool,  at  32  cents  a  pound,  and  receives  a  com- 
mission of  2f  per  cent. ;  what  amount  does  he  receive  ? 

Ans.  $235.40. 

9.  If  I  sell  400  bales  of  cotton,  each  weighing  570  pounds, 
at  9  cents  a  pound,  and  receive  a  commission  of  2J  per  cent, 
how  much  do  I  make  by  the  transaction  ?      Ans.  $461.70. 

10.  A  commission  merchant  in  New  Orleans  sells  450 
barrels  of  flour  at  $7.60  a  barrel ;  38  firkins  of  butter,  each 
containing  56  pounds,  at  25  cents  a  pound;  and  105  cheeses, 
each  weighing  48  pounds,  at  9  cents  a  pound ;  how  much 
is  his  commission  for  selling,  at  5^  per  cent.  ? 

Ans.  1242.308. 

11.  A  lawyer  collected  a  note  of  $950,  and  charged  6|-  per 
cent,  commission ;  what  was  his  fee,  and  what  the  sum  to 
be  remitted?  Ans,  Fee,  $61.75  ;  remitted,  $888.25. 

12.  An  insurance  agent's  fees  are  6  per  cent,  on  all  sums 
received  for  the  company,  and  4  per  cent,  additional  on  all 
sums  remaining,  at  the  end  of  the  year,  after  the  losses  are 
paid ;  he  receives,  during  the  year,  $30456.50,  and  pays  losses 
to  the  amount  of  $19814.15  ;  how  much  commission  does  he 
receive  during  the  year  ?  Ans.  $2253.084. 

Case  IL 

244.  To  find  the  commission  or  brokerage,  when 
it  is  to  be  dediicted  from  the  given  sum,  and  the 
balance  invested. 

1.  A  merchant  sends  his  agent  $1260  with  which  to  buy 
merchandise,  after  deducting  his  commission  of  5  per  cent ; 
what  is  the  sum  invested,  and  how  much  is  the  commission? 

OPERATION. 

$1260  -^  1.05  =  $1200,  invested. 

$1260  —  $1200  =  $60,  commiesion. 

Or,  ifj  +  T*ir  =  fi  ;   ^1260  -^  U  =  $1200,  invested , 

And  $1260  —  $1200  =  $60,  commission. 


Case  II  is  what  ?    Give  explanation.    Rule. 


COMMISSIOK    AND     BROKERAGE.  215 

Analysis.  Since  the  commission  is  5  per  cent.,  the  agent  must 
receive  $1.05  for  every  $1  he  expends  ;  he  can  invest  as  many  dollars 
as  ^1.05  is  contained  times  in  $1260,  which  is  $1200  ;  and  the  differ- 
ence between  the  given  sum  and  the  sum  invested  is  his  commission. 

Or,  the  money  expended  is  {^  of  itself,  the  commission  is  yf^  of 
this  sum,  and  the  commission  added  to  the  sum  expended  is  \^  of 
the  whole  sum.  Since  $1260  is  \^%  =  U,  $1260  -^  |^  =  $1200,  the 
sum  expended  ;  and  $1260  —  $1200  =  $60  the  commission. 

Rule.  I.  Divide  the  given  amount  by  1  increased  by  the 
rate  per  cent,  of  commission,  and  the  quotient  is  the  sum 
invested. 

II.  Subtract  tlie  investment  from  the  given  amount,  and 
the  remainder  is  the  commission. 


Examples  for  Practice. 

2.  A  man  sends  $3246.20  to  his  agent  in  Boston,  request- 
ing him  to  lay  it  out  in  shoes,  after  deducting  his  commis- 
sion of  2  per  cent. ;  what  is  his  commission  ?    Ans.  $63.65. 

3.  What  amount  of  stock  can  he  bought  for  $9682,  and 
allow  3  per  cent,  brokerage  ?  Ans.  $9400. 

4.  A  flour  merchant  sent  $10246.50  to  his  agent  in  Chi- 
cago, to  invest  in  flour,  after  deducting  his  commission  of 
3^  per  cent. ;  how  many  barrels  of  flour  could  he  buy  at 
$5.50  per  barrel?  Ans.  1800  barrels. 

5.  An  agent  receives  a  remittance  of  $4908,  with  which 
to  purchase  grain,  at  a  commission  of  4J-  per  cent. ;  what 
will  be  the  amount  of  the  purchase  ? 

6.  Remitted  $603.75  to  my  agent  in  New  York,  for  the 
purchase  of  merchandise,  agent's  commission  being  5  per 
cent. ;  what  amount  of  broadcloth  at  $5  per  yard  should  I 
receive?  Ans.  115  yds. 

7.  A  commission  merchant  receives  $9376.158,  with  or- 
ders to  purchase  grain  ;  his  commission  is  3  per  cent. ,  and 
he  charges  IJ  per  cent,  additional  for  guaranteeing  its  de- 
livery at  a  specified  time;  how  much  will  he  pay  out,  and 
what  are  hid  fees  ?  Ans.  Fees,  $403,758. 


21G  PERCE  NT  AGE. 

8.  A  real-estate  broker,  whose  stated  commission  is  1| 
per  cent.,  receives  $13842.07,  to  be  used  in  the  purchase  of 
city  lots  ;  how  much  does  he  invest,  and  what  is  his  com- 
mission?       Ans.  $13604  invested  ;  $238.07  commission. 

9.  A  broker  received  $10650,  to  be  invested  in  stocks  afte? 
deducting  J  per  cent,  for  brokerage  ;  what  amount  of  stock 
did  he  purchase  ? 

STOCK-JOBBINa. 

245.  A  Corporation  is  a  body  authorized  by  a  general 
law,  or  by  a  special  charter,  to  transact  business  as  a  single 
individual. 

246.  A  Charter  is  the  legal  act  of  incorporation,  and 
defines  the  powers  and  obligations  of  the  incorporated  body. 

24T.  A  Firm  is  the  name  under  which  an  unincorpo- 
rated company  transacts  business, 

248.  Capital  or  Stock  is  the  property  or  labor  of  an 
individual,  corporation,  company,  or  firm ;  ifc  receives  dif- 
ferent names,  as  Bank  Stock,  Kailroad  Stock,  Government 
Stock,  etc. 

249.  A  Share  is  one  of  the  equal  parts  into  which  the 
stock  is  divided. 

250.  Stockholders  arc  the  owners  of  the  shares. 

251.  The  Nominal  or  Par  Value  of  stock  is  its  first 
cost,  or  original  valuation. 

The  original  valuation  of  a  share  varies  in  different  companies.  A  share  of  bank, 
insurance,  railroad,  or  like  stock  is  usually  $100. 

252.  Stock  is  At  Par  when  it  sells  for  its  first  cost,  or 
original  valuation. 

253.  Above  Par,  at  a  premium,  or  in  advance,  when  it 
sells  for  more  than  its  original  cost ;  and 

254.  Below  Par,  or  at  a  discount,  when  it  sells  for  less 
than  its  original  cost. 

Define  a  corporation.  A  charter.  A  firm.  Capital  or  stock.  Shares. 
StQckholders.    Par  value.    At  par.    Above  par.    Below  par. 


STOCKS.  317 

255.  The  Market  or  Real  Value  of  stock  is  what  it 
will  bring  per  share  in  money. 

256.  A  Dividend  is  a  sum  paid  to  stockholders  from 
the  profits  of  the  business  of  the  company. 

357.  An  Assessment  is  a  sum  required  of  stockholders 
to  meet  the  losses  or  expenses  of  the  business  of  the  company. 

25S,  Premium  or  advance,  and  discount  on  stock,  divi- 
dends, and  assessments,  are  computed  at  a  certain  per  cent, 
upon  the  original  value  of  the  shares  of  the  stock. 

259.  A  Stock,  Broker  is  a  person  who  buys  and  sells 
stocks,  either  for  himself,  or  as  the  agent  of  another. 

260.  The  calculations  in  stock-jobbing  are  based  upon 
the  following  relations  : 

I.  Premium,  discount,  and  brokerage  are  each  a  percent- 
age, computed  upon  the  par  value  of  the  stock  as  the  base. 

II.  The  market  value  of  stock,  or  the  proceeds  of  a  sale, 
is  the  amount  or  difference,  according  as  the  sum  is  greater 
or  less  than  the  par  value. 

In  all  examples  relating  to  stocks,  $100  will  be  considered  a  share,  unless  other- 
wise stated. 

Case  I. 

261.  To  find  the  value  of  stock,  when  at  an  ad- 
vance, or  discount. 

1.  What  will  $3240  of  Bank  Stock  cost  at  8  per  cent,  ad- 
vance, brokerage  ^  per  cent.  ? 

OPERATION.  Analysis.    To  find  the 

$1  +  .08  =  $1.08.  price  of  stock,  we  add  the 

$1.08  +  .0025  =  $1.0825.  rate  of  advance  to,  or  sub- 

$1.0825   X  3240  =  $3507.30  tract  the  discount  from,  $1,- 

to  this  result  add  the  bro- 
kerage, and  we  have  the  cost  of  $1.  Hence  $3240  stock  will  cost  3240 
times  $1.0825. 

EuLE.  Multiply  the  cost  of  $1  by  the  number  indicating 
the  par  value  of  the  stock. 

Market  value.  A  dividend.  An  assessment.  Case  I  is  what  ?  Give 
explanation.     Rule. 

E.P.  10 


218  percentage. 

Examples  for  Practice. 

2.  If  the  stock  of  an  insurance  company  sell  at  5  per  cent, 
below  par,  what  will  $1200  of  the  stock  cost  ?  Ans,  $1140. 

3.  What  is  the  market  value  of  35  shares  of  New  York  Cen- 
tral Railroad  stock,  at  15  per  cent,  below  par  ?  Ans.  $2975. 

4.  What  must  be  paid  for  48  shares  of  Panama  Railroad 
stock,  at  a  premium  of  5^  per  cent.,  if  the  par  value  be  $150 
per  share,  brokerage  i  per  cent.  ?  Ans.  $7632. 

5.  What  costs  $5364  stock  in  Minnesota  copper  mines,  at 
9  per  cent,  above  par,  brokerage  ^  per  cent.  ?    Ans.  $5853.465. 

6.  A  man  purchased  $6275  stock  in  the  Pennsylvania  Coal 
Company,  at  par,  and  sold  the  same  at  a  discount  of  12  per 
cent. ;  what  was  his  loss  ?  Ans.  $753. 

•  7.  What  must  be  paid  for  125  shares  of  United  States 
stock,  at  4J  per  cent,  premium,  the  par  value  being  $1000 
per  share,  brokerage  J-  per  cent.  ?  Ans.  $131250. 

8.  Bought  42  shares  of  Illinois  Central  Railroad  stock,  at 
14  per  cent,  discount,  and  sold  the  same  at  a  premium  of 
12-J^  per  cent. ;  what  did  I  gain  ?  An^.  $1113. 

9.  What  is  the  market  value  of  175  shares  of  stock  in  the 
Suffolk  Bank,  at  f  per  cent,  advance  ?      Ans.  $17631.25. 

10.  Bought  75  shares  of  stock  in  the  Bank  of  New  Orleans 
at  $50  each,  at  3  per  cent,  discount,  and  sold  it  at  2J  per 
cent,  advance ;  what  was  my  gain  ?  Ans.  $196,875. 

11.  B.  exchanged  28  shares  of  bank  stock,  of  $50  each, 
worth  7  per  cent,  premium,  for  25  shares  of  railroad  stock,  of 
$100  each,  at  12^  per  cent,  discount,  and  paid  the  difference 
in  cash  ;  how  much  cash  did  he  pay?  Ans.  $689.50. 

12.  A  speculator  exchanged  $3600  of  Railroad  bonds,  at 
5  per  cent,  discount,  for  27  shares  of  Bank  stock,  at  3  per 
cent,  premium,  receiving  the  difference  in  cash  ;  how  much 
money  did  he  receive  ?  Ans.  $639. 

13.  I  bought  120  shares  Pacific  Railroad  stock,  at  a  dis- 
count of  2J  percent.,  and  sold  the  same  at  an  advance  of  12 
per  cent.;  what  was  my  gain?  Afis.  $1740. 


STOCKS.  219 

Case  II. 
362.  To  find  how  much  stock  may  be  purchased 
for  a  given  sum. 

1.  How  many  shares  of  bank  stock,  at  3  per  cent,  advance, 
may  be  bought  for  $5150  ? 

OPERATION.  Analysis.    Since  the  stock 

$5150  -^  1.03  =  15000  =  is  at  3  per  cent,  advance,  $1 

50  shares,  Ans.  ^^  s*«^^  ^*  P^^  ^^'^^^  "^^^  ^^-^^  - 

and  if  we  divide  $5150,  the 

whole  sum  to  be  expended,  by  1.08,  the  cost  of  $1  of  stock,  the  quo- 
tient must  be  the  amount  of  stock  purchased. 

Rule.  Divide  the  given  sum  hy  tJie  cost  of  $1  of  stoch,  and 
the  quotie7it  will  he  the  nominal  amount  of  stock  purchased. 

2.  How  many  shares  of  railroad  stock,  at  5  per  cent,  ad- 
vance, can  be  purchased  for  $6300  ?  Ans.  GO  shares. 

3.  I  invested  $6187.50,  in  Ocean  Telegraph  stock,  at  10 
per  cent,  discount ;  how  much  stock  did  I  purchase  ? 

Ans.  S6875. 

4.  I  sent  my  agent  $53500  to  be  invested  in  Hlinois  Cen- 
tral Railroad  stock,  which  was  selling  at  7  per  cent,  advance; 
what  amount  did  he  purchase  ?  Ans.  $50000. 

5.  Sold  50  shares  of  stock  in  a  Pittsburgh  ferry  company, 
at  8  per  cent,  discount,  and  received  $1150 ;  what  is  the  par 
value  of  1  share  ?  Ans.  $25, 

Stock  Intestmen'ts. 
263.  The  net  earnings  of  a  corporation  are  usually  divid- 
ed among  the  stockholders,  in  semi-annual  dividends.  The 
income  of  capital  stocJc  is  therefore  fluctuating,  being  depend- 
ent upon  the  condition  of  business ;  while  the  income  arising 
from  hands,  whether  of  government  or  corporations,  is  fixed, 
being  a  certain  rate  per  cent.,  annually,  of  the  par  value,  or 
face  of  the  bonds. 

Case  II  is  what?  Give  explanation.  Explain  difference  between 
income  of  capital  stock,  and  of  bonds. 


220  PERCEN^TAGE. 

264.  Federal  or  United  States  Securities  are  of  two 

kinds;  viz.,  Bonds  and  Notes. 

Bonds  are  of  two  kinds. 

First,  Those  which  are  payable  at  a  fixed  date,  and  are 
known  and  quoted  in  commercial  transactions  by  the  rate 
of  interest  they  bear,  thus  :  U.  S.  6's,  that  is,  United  States 
Bonds  bearing  Q>%  interest. 

Second,  Those  which  are  payable  at  a  fixed  date,  but 
which  may  be  paid  at  an  earlier  specified  time,  as  the  Gov- 
ernment may  elect.  These  are  known  and  quoted  in  com- 
mercial transactions  by  a  combination  of  the  two  dates, 
thus :  U.  S.  5-20's ;  or  a  combination  of  the  rate  of  inter- 
est and  the  two  dates,  thus  :  U.  S.  6's  5-20  ;  that  is,  bonds 
bearing  Q%  interest,  which  are  payable  in  twenty  years,  but 
may  be  paid  in  five  years,  if  the  Government  so  elect. 

When  it  is  necessary,  in  any  transaction,  to  distinguish 
from  each  other  different  issues  which  bear  the  same  rate  of 
interest,  this  is  done  by  adding  the  year  in  which  they  be- 
come due,  thus  :  U.  S.'5's  of  71  ;  U.  S.  5's  of  '74  ;  U.  S.  6'i? 
5-20  of  '84;  U.  S.  6's  5-20  of  '85. 

Notes  are  of  two  kinds. 

First,  Those  payable  on  demand,  without  interest,  known 
as  United  States  Legal-tender  Notes,  or,  in  common  lan- 
guage, *^  Greenbacks." 

Second,  Notes  payable  at  a  specified  time,  with  interest, 
known  as  Treasury  Notes.  Of  these  there  are  two  kinds, — 
Q%  Compound-interest  Notes,  and  Notes  bearing  ^7^%  in- 
terest, the  latter  known  and  quoted  in  commercial  transac- 
tions as  7.30's. 

The  nomenclature  here  explained  is  that  used  in  com- 
mercial transactions,  which  involve  similar  Securities  of 
States  or  Corporations. 


What  are  United  States  Securities  composed  of  ?  Explain  the  dif- 
ferent kinds  of  bonds.  Of  Notes.  In  what  is  the  interest  on  each 
payable  ? 


STOCKS.  221 

The  interest  on  all  U.  S.  bonds  is  payable  in  gold. 
The  interest  on  notes  is  payable  in  Legal- tender  Notes. 
The  following  are  the  principal  United  States  Securities : 
Bonds. 


U.  S.  6's  of  1867. 
U.  S.  6's  of  1868. 
U.  S.  6's  of  1880. 
U.  S.  6's  of  1881. 
U.  S.  5's  of  1871. 
U.  S.  5's  of  1874. 
U.  S.  5's  (New)  of  1881. 
U.  S.  4i's    "      of  1886. 


U.  S.  4's  (New)  of  1901. 
U.  S.  5-20's,  due  in  1882,  interest  Qfo, 
U.  S.  5-20's,  due  in  1884,  interest  6%. 
U.  S.  5-20's,  due  in  1885,  interest  6%. 
U.  S.  10-40's,  due  in  1904,  interest  5%. 
Pacific  Railroad  6's  of  1895. 
Pacific  Railroad  6's  of  1896. 


Notes, 
Compound-interest  Notes  of  1867.  I        7.30  Notes  of  1867. 

Compound-interest  Notes  of  1808.  I        7.30  Notes  of  1 868. 

1.  The  5-20's  were  issued  in  1862,  '61,  '65,  '67,  and  '70.  They  bear 
interest  at  6^,  paid  semi-annually  in  gold,  except  the  issue  of  1870, 
called  5's  of  '81,  which  bear  interest  at  5%,  paid  quarterly  in  gold. 

2.  Bonds  issued  by  States,  cities,  etc.,  are  quoted  in  a  similar  man- 
ner. Thus,  S.  G.  6's  are  bonds  bearing  6%  interest,  issued  by  the 
State  of  South  Carolina. 

Case  I. 

265.  To  find  what  income  any  investment  will 
produce. 

1.  What  income  will  he  obtained  by  investing  $6840  in 
stock  bearing  6^,  and  purchased  at  95^  ? 

OPERATION.  Analysis.  We 

$6840  -^  .95  =  $7200,  stock  purchased,      divide  the  invest- 

$7200  X  .06  =  $432,  annual  income.  ^^^t'  ^^840,  by 

the  cost  of  $1,  and 
obtain  $7200,  the  stock  which  the  investment  will  purchase  (262). 
And  since  the  stock  bears  6%  interest,  we  have  $7200  x  .06  =  $432, 
the  annual  income  obtained  by  the  investment. 

Rule.  Find  hoio  much  stock  the  investment  toill  purchase, 
and  then  compute  the  income  at  the  given  rate  upon  the  par 
value. 

Name  the  different  kinds  of  bonds.  Of  Notes.  What  is  Case  I  ? 
Explanation  ?    Rule  ? 


222  PE  EC  EXT  AGE. 

2.  If  I  invest  $SG7  in  U.  S.  5-20's  of  '84  at  102^,  what 
income  will  I  receive  on  my  investment  ?  Ans.  $51. 

3.  What  will  be  my  yearly  income^  if  I  invest  $8428  in 
U.  S.  10-40's,  at  2S%  ?  Ans.  $430. 

4.  How  much  stock  at  a  premium  of  4^%  can  be  bought 
for  $10500,  brokerage  i%  ?  A7is.  $10000. 

5.  If  A  invest  $4795  in  Maryland  5's  at  87^,  brokerage 
i%,  what  will  be  his  yearly  income  ?  A7is.  $274. 

6.  Having  $10476  to  invest,  I  find  I  can  purchase  U.  S. 
6's  at  107i^,  and  U.  S.  5-20's  of  '81  at  96-^^,  brokerage  i%, 
in  each  instance.  How  much  more  will  I  receive  yearly  by 
investing  in  the  former  than  in  the  latter?         Ans.  $42. 

7.  A  having  a  farm  of  109  acres,  which  rents  for  $681.25, 
sells  the  same  for  $125  per  acre,  and  invests  the  proceeds  in 
U.  S.  6's  at  108}^,  brokerage  i%  for  purchasing ;  will  his 
yearly  income  be  increased  or  diminished,  and  how  much  ? 

Ans.  Increased  $68.75. 


Case  II. 

266.  To  find  what  sum  must  be  invested  to  ob- 
tain a  given  income. 

1.  What  sum  must  be  invested  in  Virginia  6%  bonds, 
purchasable  at  80^,  to  obtain  an  income  of  $600  ? 

OPERATION.  Analysis.— 

$600  -T-  .05  =  $12000,  stock  required.         Since  $1  of  the 

$12000  X  .80  =  $9600,  cost  or  investment,    stock  will  ob- 

tain  $.05  in- 
come, to  obtain  $600  will  require  $600 h- .05 =$12000  (Case  I).  Multi- 
plying the  par  value  of  the  stock  by  the  market  price  of  $1,  we  have 
$12000  X  .80= $9600,  the  cost  of  the  required  stock,  or  the  sum  to  he 
invested. 

Rule.  I.  Divide  the  given  income  by  the  per  cent,  which  the 
stock  pays ;  the  quotient  will  be  the  par  value  of  the  stock 
required. 

What  is  Case  II  ?    Explanation  ?    Rule  ? 


STOCKS.  223 

11.  Multiply  the  par  value  of  the  dock  hy  the  market  value 
of  one  dollar  of  the  stock  ;  the  product  zvill  be  the  required 
investment. 

Examples  for  Practice. 

2.  If  K  Y.  6's  are  6%  below  par,  wliat  sum  must  be  in- 
vested in  this  stock  to  obtain  an  income  of  $840  ? 

Ans.  $13300. 

3.  What  sum  must  be  invested  in  U.  S.  10-40's  at  98J^, 
brokerage  i%  for  buying,  to  secure  an  annual  income  of 
$1860  ?  Ans.  136642. 

4.  When  U.  S.  5-20's  of  '81  are  quoted  at  lOSJ,  what  sum 
must  I  invest  to  secure  an  annual  income  of  $1080,  broker- 
age i^?  Ans.  123436. 

5.  If  I  sell  $25000  U.  S.  5-20's  of  '82  at  93|^,  and  invest 
a  sufficient  amount  of  the  proceeds  in  U.  S.  6's,  at  109J^  to 
yield  an  annual  income  of  $960,  and  buy  a  house  with  the 
remainder,  how  much  will  the  house  cost  me  ? 

A71S.  $5957.50. 
Case  III. 

267.  To  find  what  per  cent,  the  income  is  of  the 
investment,  when  stock  is  purchased  at  a  given 
price. 

1.  What  per  cent,  of  my  investment  shall  I  secure  by 
purchasing  the  New  York  7's  at  105^  ? 

Analysis.     Since  $1  of  the  stock 
OPEKATION.  will  cost  $1.05,  and  pay  $.07,  the  in- 

.07  ^  1.05  =  6f  ^.  come  is  j^j  =  6|%  of  the  invest- 

ment. 

EuLE.  Divide  the  annual  rate  of  income  which  the  stock 
dears  hy  the  price  of  the  stock  ;  the  quotient  will  he  the  rate 
upon  the  investment. 

Examples  for  Practice. 

2.  What  is  the  rate  of  income  upon  money  invested  in 
Q%  bonds,  purchased  at  87  per  cent.  ?  Ans.  6f| ^. 

What  is  Case  III  ?    Explanation  ?    Rule  ? 


324  PERCENTAGE. 

3.  What  per  eent.  on  liis  money  will  a  man  receive  an- 
nually if  he  invest  in  N.  Y.  G's  at  105^  ?  Ans.  5\%. 

4.  What  is  the  rate  of  income  upon  money  invested  in 
Missouri  6's  at  76%  ?  Ans.  S%. 

5.  Purchased  U.  S.  5-20's  of  '84  at  107|^,  brokerage  }^; 
what  is  the  income  on  the  investment?  Ans.  5|^. 

6.  Which  is  the  better  investment,  U.  S.  10-40's  at  98J-^, 
or  U.  S.  5-20's  of  '85  at  lOSf^,  brokerage  i%  in  each? 

Case  IV. 

268.  To  find  the  price  at  which  stock  must  be 
purchased  to  obtain  a  given  rate  upon  the  invest- 
ment. 

1.  At  what  price  must  6%  stocks  be  purchased  in  order 
to  obtain  8%  income  on  the  investment  ? 

Analysis.    Since  $.06,  the  income 
OPERATION.  of  $1  of  the  stock,  is  8%  of  the  sum 

$.06  -^  .08  =  1.75  paid  for  it,  we  have  (235)  $.06^ 

.08  =  75%,  the  purchase  price. 

KuLE.  Divide  the  annual  rate  of  income  which  the  stock 
hears  hy  the  rate  required  on  the  investment ;  the  quotient 
will  he  the  price  of  the  stock. 

Examples  foe  Practice. 

2.  What  must  I  pay  for  Missouri  6's  that  my  investment 
may  yield  9^  annually  ?  Ans.  66J^. 

3.  What  rate  of  premium  does  6^  stock  bear  in  the  mar- 
ket when  an  investment  pays  h%  ?  Ans.  %0%. 

4.  At  what  rate  must  I  buy  U.  S.  10-40's  that  I  may  re- 
ceive Q%  on  my  investment?  Ans,  83-J-^. 

5.  At  what  rate  of  discount  must  U.  S.  5-20's  of  '81  be 
purchased  that  I  may  secure  7^  annually  on  the  invest- 
ment ?  Ans.  284^. 

What  is  Case  IV  ?    Explanation  ?    Rule  ? 


STOCKS.  225 

GOLD    INVESTMENTS. 

269,  Currency  is  a  term  used  in  commercial  language. 
First,  To  denote  the  aggregate  of  Specie  and  Bills  of  Ex- 
change, Bank  Bills,  Treasury  Notes,  and  other  substitutes 
for  money  employed  in  buying,  selling,  and  carrying  on  ex- 
change of  commodities  between  various  nations.  Second, 
To  denote  whatever  circulating  medium  is  used  in  any 
country  as  a  substitute  for  the  government  standard.  In 
this  latter  sense,  the  paper  circulating  medium,  when  below 
par,  is  called  Currency,  to  distinguish  it  from  gold  and 
silver.  If,  from  any  cause,  the  paper  medium  depreciates 
in  value,  as  it  has  done  in  the  United  States,  gold  becomes 
an  object  of  investment,  the  same  as  stocks.  In  commercial 
language,  gold  is  represented  as  rising  and  falling  ;  but  gold 
being  the  standard  of  value,  it  can  not  vary.  The  variation 
is  in  the  medium  of  circulation  substituted  for  gold ;  hence, 
when  gold  is  said  to  be  at  a  premium,  the  currency,  or  cir- 
culating medium,  is  made  the  standard,  while  it  is  Adrtually 
below  par. 

Case  I. 

370.  To  change  gold  into  currency. 

1.  How  much  currency  can  be  bought  for  1150  in  gold 
when  gold  is  at  170^  ? 

OPERATION.  Analysis.    Since  a  dollar  of  gold 

$1.70  X  150  =  $255.  is  worth  $1.70  in  currency,  there  can 

be  as  many  times  $1.70  of  currency- 
bought  as  there  are  dollars  of  gold.  Therefore,  $1.70  x  150  =  $255 
^a  the  amount  of  currency  which  can  be  purchased  for  $150  in  gold. 

KuLE.  Multiply  the  value  of  one  dollar  of  gold  in  currency 
hy  the  number  of  dollars  of  gold. 

2.  What  is  the  value  in  current  funds  of  $250  gold,  when 
gold  is  147^  ?  Ans,  $367.50. 

3.  What  is  the  value  in  current  funds  of  $320.50,  when 
goldisl37i^?  Ans.  $440.68J. 


What  is  Currency?    Case  I?    Explanation?    Rule? 
10* 


226  PEKCEKTAGE. 

4.  When  gold  is  at  a  premium  of  33%,  how  much  will 
$2500  in  gold  cost?  Ans,  $3325. 

5.  A  holds  $8000  U.  S.  10-40's  ;  what  is  his  annual  income 
in  currency  if  gold  is  138  ?  Ajis.  $552. 

6.  What  is  the  yearly  income  in  currency  from  $9500  of 
U.  S.  5-20's  of  '84  when  gold  is  140  ?  Ans,  $798. 

7.  A  purchased  a  house,  for  which  he  was  to  pay  $450C 
in  currency,  or  $3000  in  gold  at  his  option.  Will  he  gain 
or  lose  by  accepting  the  latter  offer,  gold  being  147^%,  and 
how  much  in  currency?  Ans,  Gain  $75. 

Case  II. 
271.   To  change  currency  into  gold. 

1.  How  much  gold  can  be  purchased  for  $75  current 
funds,  gold  being  at  150^  ? 

AifALYSis.    A  dollar  of  gold  cost  $1.50 

OPEBATION.  in  currency,  therefore  there   can  be  as 

$75  _i_  $1.50  =  50.        many  dollars  of  gold  purchased  for  $75  in 

currency  as  $1.50  is  contained  times  in  $75. 

Eule.  Divide  the  amount  in  currency  by  the  price  of  gold, 

2.  What  is  the  value  in  gold  of  a  dollar  in  currency,  when 
gold  is  quoted  at  138^^  ?  A7is,  $72^^^. 

3.  Gold  being  the  standard,  what  is  the  rate  of  discount 
upon  current  funds,  when  gold  is  at  145,  147,  195|-,  280^  ? 

Ans.  to  last,  Q4^%, 

4.  How  much  gold  can  be  purchased  for  $4181  current 
funds,  when  gold  is  quoted  at  148^?  Ans,  $2825. 

5.  If  I  sell  prints  for  24  cents  per  yard,  in  currency,  what 
is  the  price  in  gold,  gold  being  at  160^  ?    Ans.  15  cents. 

6.  Sold  $5900  U.  S.  10-40's  at  90^,  and  invested  the 
proceeds  in  gold  at  147J  ;  how  much  gold  did  I  purchase  ? 

Ans.  $3G00. 

7.  What  is  the  value  in  gold  of  a  dollar,  in  currency, 
when  gold  is  at  145^  ?  Ans.  $.68f  f. 

8.  I  invested  $792  of  currency  in  gold,  when  gold  is  quoted 
at  165^.     How  much  gold  did  I  purchase  ?     Ans,  $480. 

What  is  Case  II  ?    Explanation  1    Rule  t 


PEOFIT    AND    LOSS.  227 

9.  What  is  gold  quoted  at,  when  a  dollar  in  currency  is 
worth  30  cents  in  gold  ?  45  cts.  ?  54  cts.  ?  60  cts.  ?  74  cts.  ? 

10.  How  many  yards  of  cloth  at  $3.50  in  gold  can  be 
bought  for  $126,  currency,  when  gold  is  at  140  ?    Ans.  25^. 

11.  Bought  flour  at  $11.75  per  barrel  in  currency,  when 
gold  was  at  150^,  and  afterwards  sold  it  at  $10.35  in  cur- 
rency, when  gold  was  135^ ;  did  I  gain  or  lose,  and  how 
much,  on  a  sale  of  300  barrels  ?  Ans.  $50. 

12.  Which  is  the  better  investment,  a  bond  and  mortgage 
at  7%,  or  U.  S.  5-20's  of  '84  at  par,  gold  being  140^  ;  and 
what  per  cent,  in  gold?  Ans.  XJ.  S.  5-20's  1%. 

13.  Sold  $51100  7-30  Treasury-Notes,  at  104^,  and  in- 
vested the  proceeds  in  gold  at  146^,  with  which  I  bought 
U.  S.  10-40's  at  70^  in  gold.  Will  my  yearly  income  be 
increased,  or  diminished  by  the  transaction,  and  how  much 
in  gold  ?  Ans.  Increased  $45. 

PROFIT    AND    LOSS. 

272.  Profit  and  Loss  are  commercial  terms,  used  to 
express  the  gain  or  loss  in  business  transactions,  which  is 
usually  reckoned  at  a  certain  per  cent,  on  the  prime  or  first 
cost  of  articles. 

Case  I. 

273.  To  find  the  amoiint  of  profit  or  loss,  when 
the  cost  and  the  gain  or  loss  per  cent,  are  given. 

1.  A  man  bought  a  horse  for  $135,  and  afterward  sold 
him  for  20^  more  than  he  gave  ;  how  much  did  he  gain  ? 

OPERATION.  Analysis.    Since  $1 

$135  X  .20  =  $27,   Ans,  gains  20  cents,  or  20%, 

Or,  A'lr  =  -J  ;  ^135  x  i  =  $27.  $1^5  will  gain  $135  x  .20 

=  $27.    Or,  since  20% 
equals  yV^  =  I,  the  whole  gain  will  be  ^  of  the  cost. 

Rule.  Multiply  the  cost  hy  the  rate  per  cent,  expressed 
decimally.    Or, 

Take  such  part  of  the  cost  as  the  rate  per  cent,  is  part  of  100. 

What  is  meant  by  profit  and  loss  ?  Case  I.  is  what  ?  Give  explana- 
tion.    Rule. 


228  percentage. 

Examples  for  Practice. 

2.  A  grocer  bought  a  hogshead  of  sugar  for  S84.80,  and 
sold  it  at  12 J  per  cent,  profit ;  what  was  his  gain  ? 

3.  A  miller  bought  500  bushels  of  wheat  at  $1.15  a  bushel, 
and  he  sold  the  flour  at  16|  per  cent,  advance  on  the  cost 
of  the  wheat ;  what  was  his  gain?  Ans.  $95.83^. 

4.  Bought  76  cords  of  wood  at  13.62-J-  a  cord,  and  sold  it 
so  as  to  gain  26  per  cent. ;  what  did  I  make  ? 

5.  A  hatter  bought  40  hats  at  $1.75  apiece,  and  sold  them 
at  a  loss  of  14^  per  cent.  ;  what  was  his  whole  loss  ? 

6.  A  grocer  bought  3  barrels  of  sugar,  each  containing  230 
pounds,  at  8}  cents  a  pound,  and  sold  it  at  18^^  per  cent, 
profit ;  what  was  his  whole  gain,  and  what  the  selling  price 
per  pound?     Ans.  Gain,  $10.35  ;  price  per  lb.,  9|  cents. 

7.  A  sloop,  freighted  with  3840  bushels  of  corn,  encoun- 
tered a  storm,  when  it  was  found  necessary  to  throw  37 J  per 
cent,  of  her  cargo  overboard  ;  what  was  the  loss,  at  62|-  cents 
a  bushel  ?  Ans.  $900  loss. 

8.  A  gentleman  bought  a  store  and  contents  for  $4720  ; 
he  sold  the  same  for  12^  per  cent,  less  than  he  gave,  and 
then  lost  15  per  cent,  of  the  selling  price  in  bad  debts  ;  what 
was  his  entire  loss  ?  A7is.  $1209.50. 

9.  A  man  commenced  business  with  $3000  capital ;  the 
first  year  he  gained  22^  per  cent,  which  he  added  to  his 
capital ;  the  second  year  he  gained  30  per  cent,  on  the  whole 
sum,  which  gain  he  also  put  into  his  business ;  the  third 
year  he  lost  16|  per  cent,  of  his  entire  capital ;  how  much 
did  he  make  in  the  3  years  ?  Ans.  $981.25. 

Case  II. 

274.  To  find  the  gain  or  loss  per  cent.,  when  the 
cost  and  selling  price  are  given. 

1.  Bought  wool  at  32  cents  a  pound,  and  sold  it  for 
40  cents  a  pound  ;  what  per  cent,  was  gained  ? 

Case  II  is  what  ?     Givo  explanation.     Rule, 


PROFIT     AXD     LOSS.  229 

OPERATION. 

40  —  32  =  8  ;  8  -^  32  =  ^  =  .25,  Ans. 
Or,  40  -  32  =  8  ;  8  -^  32  =  -^  =  J ;  i  X  100  =  25^. 

Analysis.  Since  the  gain  on  32  cents  is  40  —  33  =  8  cents,  the 
whole  gain  is  -i^  =  i  of  the  purchase  money  ;  and  ^  reduced  to  a 
decimal  is  25  hundredths,  equal  to  25  per  cent.  Or,  if  the  gain  were 
equal  to  the  purchase  money,  it  would  be  100  per  cent.  ;  but  since  the 
gain  is  ^  =  ^  of  the  purchase  money,  it  will  be  ^  of  100  per  cent., 
equal  to  25  per  cent. 

Rule.  Make  the  difference  betioeen  the  purchase  andseUi7ig 
prices  the  numerator,  and  the  purchase  price  the  denomiiia- 
tor  ;  reduce  to  a  decimal,  and  the  result  luill  he  the  per  cent. 

Or,  Tahe  such  a  part  of  100  as  the  gain  or  loss  is  part  of 
the  purchase  price. 

Examples  for  Practice. 

2.  A  man  oought  a  pair  of  horses  for  1275,  and  sold  them 
for  $330  ;  what  per  cent,  did  he  gain  ?  Ans.  20^. 

3.  If  a  merchant  buy  cloth  at  $.60  a  yard,  and  sell  it  for 
$.75  a  yard,  what  does  he  gain  per  cent.  ? 

4.  A  speculator  bought  108  barrels  of  flour  at  $4.62|-  a 
barrel,  and  sold  it  so  as  to  gain  $114.88|- ;  what  per  cent, 
profit  did  he  make  ?  Ans.  23  per  cent. 

5.  Bought  sugar  at  8  cents  a  pound,  and  sold  it  for 
9J  cents  a  pound  ;  what  per  cent,  was  gained  ? 

6.  A  drover  bought  150  head  of  cattle  for  $42  per  kead, 
and  sold  them  for  $5400  ;  what  was  his  loss  per  cent,  i 

Ans.  14^%. 

7.  If  I  sell  for  $15  what  cost  me  $25,  what  do  I  lose  per 
cent.  ?  Ans.  40  per  cent. 

8.  Bought  paper  at  $2  per  ream,  and  sold  it  at  25  cents 
a  quire  ;  what  was  the  gain  ?  Ans.  150^. 

9.  If  I  sell  ^  of  an  article  for  f  of  its  cost,  what  is  gained 
per  cent.  ?  Ans.  50  per  cent. 

10.  If  f  of  an  article  be  sold  for  what  ^  of  it  cost,  what  is 
the  loss  per  cent.  ?  Ans.  37 J  per  cent. 


230  PERCENTAGE. 

11.  If  I  sell  3  pecks  of  clover-seed  for  what  one  bushel 
cost  me,  what  per  cent,  do  I  gain  ?  Ans.  33^%. 

12.  A,  having  a  debt  against  B,  agreed  to  take  $.87J^  on 
the  dollar  ;  what  per  cent,  did  A  lose  ? 

13.  A  grocer  bought  7  cwt.  20  lb.  of  sugar,  at  7  cents  a 
pound,  and  sold  3  cwt.  42  lb.  at  8  cents,  and  the  remainder 
at  8-J-  cents ;  what  was  his  gain  per  cent.  ?  Ans.  18^  per  cent. 

14.  Bought  2  hogsheads  of  wine,  at  $1.25  a  gallon,  and 
sold  the  same  at  $1.60  ;  what  was  the  whole  gain,  and  what 
the  gain  per  cent  ?  Ans.  Gain  28^. 

15.  A  grain  dealer  bought  com  at  $.55  a  bushel  and  sold 
it  at  $.66,  and  wheat  for  $1.10,  and  sold  it  for  $1.3 7^- ;  upon 
which  did  he  make  the  greater  per  cent.  ? 

Ans,  5  per  cent.,  upon  the  wheat. 


Case  III. 

275.  To  find  the  selling  price,  when  the  cost  and 
the  gain  or  loss  per  cent,  are  given. 

1.  Bought  a  horse  for  $136 ;  for  how  much  must  he  be 
sold  to  gain  25  per  cent.  ? 

OPERATION.  Analysts.    Since  $1  of  cost 

$1  +  .  25  =  $1.25.  sells  for  $1.25,  $1 3G  of  cost  will 

$1.25  X  136  =  $170,  Ans.       ^^^^  *'**^  1^^  *^^^®s  ^^■^^'  ^^^^^ 

Or  XM.  J-    26    —  12  5  _  8  ®<l^^^s  $170,  the  selling  price. 

^h  W  1-  Tow  —  ttrtr  —  Z'  Or,  since  the  cost  is  {U,  and 

$136  X  I  =  $170,  Ans,  ^j^g  ^^^  ^s^p^^  the  selling  price 

will  be  \l^  =  f  of  the  cost,  or 
f  of  $136  =  $170.  If  the  horse  had  been  sold  at  a  loss  of  25  per  cent., 
then  $1  of  cost  would  have  sold  for  $1  minus  .25,  or  $.75,  etc. 

Rule.  Multiply  $1  increased  ly  the  gain  or  diminished  ly 
the  loss  per  cent  hy  the  nuraher  denoting  the  cost.     Or, 

Talce  such  a  part  of  the  cost  as  is  equal  to  |^g-  increased  or 
diminished  ly  the  gain  or  loss  per  cent. 

Case  III  is  what  ?     Give  explanation.    Role. 


profit   and   loss.  231 

Examples  for  Practice. 

2.  If  12^  hundred-weight  of  sugar  cost  1140,  how  must 
it  be  sold  per  pound  to  gain  26%  Y  Ans.  14  cents. 

3.  Bought  a  hogshead  of  molasses  for  30  cents  a  gallon, 
and  paid  16f  per  cent,  on  the  prime  cost,  for  freight  and 
cartage  ;  what  must  it  sell  for,  per  gallon,  to  gain  33^  per 
cent,  on  the  whole  cost  ?  A^is.  $.46f . 

4.  For  what  price  must  I  sell  coffee  that  cost  10^  cents  a 
pound,  to  gain  17-J-^  ? 

5.  If  I  am  compelled  to  sell  damaged  goods  at  a  loss  of 
15  per  cent,  how  should  I  mark  goods  that  cost  me  $.62J-? 
11.20?  I3.87i^?  A71S.  $.53i;  $1.02;  $3.29|. 

6.  A  man,  wishing  to  raise  some  money,  offers  his  house 
and  lot,  which  cost  him  $3240,  for  18  per  cent,  less  than 
cost ;  what  is  the  price  ? 

7.  C  bought  a  farm  of  120  acres,  at  $28  an  acre,  paid 
$480  for  fencing,  and  then  sold  it  for  12^^  per  cent,  advance 
on  the  whole  cost ;  what  was  his  whole  gain,  and  what  did 
he  receive  an  acre  ?  Ans.  $480  gain  ;  $36  an  acre. 

8.  Bought  a  cask  of  brandy,  containing  52  gallons,  at 
$2.60  per  gallon  ;  if  7  gallons  leak  out,  how  must  the  re- 
tiiainder  be  sold  per  gallon,  to  gain  87^-  per  cent,  on  the 
cost  of  the  whole  ?  Ans,  $4.13}. 

9.  A  merchant  bought  15  pieces  of  broadcloth,  each  piece 
containing  23^  yards,  for  $840,  and  sold  it  so  as  to  gain 
18|  per  cent. ;  what  did  he  receive  a  yard? 

Case  IV. 

276.  To  find  the  cost,  when  the  selling  price  and 
the  gain  or  loss  per  cent,  are  given. 

1.  A  merchant  sold  cloth  for  $4.80  a  yard,  and  by  so  doing 
made  33-J  per  cent. ;  how  much  did  it  cost  ? 

OPERATIOJT. 

1  +  .33i  =  1.33J  ;  $4.80  -^  1.33^  =  $3.60,  A7is. 
Or,  $4.80  =  f  of  the  cost ;  $4.80  -r-  i  =  $3.60. 

Case  IV  is  what? 


PERCENTAGE. 

Analysis.  Since  the  gain  is  83^  per  cent,  of  the  cost,  $1  of  the 
cost,  increased  by  33^  per  cent.,  will  be  what  $1  of  cost  sold  for ; 
therefore  there  will  be  as  many  dollars  of  cost,  as  1.33^  is  contained 
times  in  $4.80,  or  $3.60.  Or,  since  he  gained  33}  per  cent.  =  ^  of  the 
cost,  $4.80  is  I  of  the  cost ;  $4.80  -i-  |  =  $3.60. 

If  the  rate  per  cent,  be  loss,  subtract  it  from  1,  instead  of  adding  it. 

Rule.  Divide  the  selling  price  hy  1  increased  hy  the  gain 
or  diyninished  ly  the  loss  per  cent.,  expressed  decimally,  or  in 
the  form  of  a  common  fraction,  and  the  quotient  will  le  the 
cost. 

Examples  for  Practice. 

2.  By  selling  sugar  at  8  cents  a  pound,  a  merchant  lost  20 
per  cent. ;  what  did  the  sugar  cost  him?    Ans.  10  cents. 

3.  Sold  flour  for  $6.12|^  per  barrel,  and  lost  12|  per  cent. ; 
what  was  the  cost  ?  Ans.  $7.00. 

4.  A  grocer,  by  selling  tea  at  $.96  a  pound,  gains  28  per 
cent. ;  what  did  it  cost  him  ?  Ans.  $.75. 

5.  Sold  a  quantity  of  flour  for  $1881,  which  was  18}  per 
cent,  more  than  it  cost :  what  did  it  cost  ? 

6.  Sold  25  barrels  of  apples  for  $69. 75,  and  made  24  per 
cent. ;  what  did  they  cost  per  barrel  ? 

7.  Sold  9^  cwt.  of  sugar  at  $8:1-  per  cwt.,  and  thereby  lost 
12  per  cent. ;  what  was  the  whole  cost  ? 

8.  Having  used  a  carriage  six  months,  I  sold  it  for  $96, 
which  was  20  per  cent,  below  cost ;  what  would  I  have  re- 
ceived had  I  sold  it  for  15  per  cent,  above  cost  ?    Ans.  $138. 

9.  B  sells  a  pair  of  horses  to  C,  and  gains  1%^  per  cent. ; 
C  sells  them  to  D  for  $570,  and  by  so  doing  gains  18|  per 
cent. ;  what  did  the  horses  cost  B?  Ans.  $426.66|. 

10.  A  grocer  sold  4  barrels  of  sugar  for  $24  each ;  on  2 
barrels  he  gained  20  per  cent.,  and  on  the  other  2  he  lost  20 
per  cent. ;  did  he  gain  or  lose  on  the  whole  ?    Ans.  Lost  $4, 

11.  A  person  sold  out  his  interest  in  business  for  $4900, 
which  was  40  per  cent,  more  than  3  times  as  much  as  he  be- 
gan ^vith  ;  how  much  did  he  begin  with  ?    Ans.  $1166.66J. 

Give  explanation.    Rule. 


IKSURAIS^CE.  233 


INSUEANCE. 

277.  Insurance  on  property  is  security  guaranteed  by 
one  party  to  another,  for  a  stipulated  sum,  against  the  loss 
of  that  property  by  fire,  navigation,  or  any  other  casualty. 

278.  The  Insurer  or  Underwriter  is  the  party  taking 
the  risk. 

279.  The  Insured  is  the  party  protected. 

280.  The  Policy  is  the  written  contract  between  the 
parties. 

281 .  The  Premium  is  the  sum  paid  by  the  insured  to 
the  insurer,  and  is  estimated  at  a  certain  rate  per  cent,  of 
the  amount  insured,  which  rate  yaries  according  to  the 
degree  of  hazard,  or  class  of  risk. 

As  a  security  against  fraud,  most  insurance  companies  take  risks  at  not  more 
than  two-thirds  the  ftdl  value  of  the  property  insured. 

282.  To  find  the  premium  when  the  rate  of  in- 
surance and  the  amount  insured  are  given. 

1.  What  must  I  pay  annually  for  insuring  my  house  to 
the  amount  of  $3250,  at  IJ  per  cent,  premium  ? 

OPEKATiON.  Analysis.    Multiply 

$3250  X  .OIJ  or  .0125  =  140.625.  the    amount    insured. 

Or,  IJ  per  ct.  =  jf^  =  ^  ;  $3250,  by  the  rate,  1^ 

$3250  X  A  =  $40,624.  P®^  *^®^*'  """^  *^'^  ^^- 

®^  ^  suit,    $40,635,    is    the 

premium.    Or,  the  rate,  1\  per  cent.,  is  -^^  =  ^V  ^^  the  amount  in- 
sured, and  i^  of  $3250  is  $40.62 1. 

Rule.  Multiply  the  amount  insured  by  the  rate  per  cent, 
and  the  product  will  he  the  premium.     Or, 

Take  such  a  part  of  the  amomit  insured  as  the  rate  is  part 
of  100. 

Define  insurance.  Insurer,  or  underwriter.  Policy.  Premium. 
To  what  amount  can  property  usually  be  insured  ?  Give  analysLs  of 
example  1.     Rule, 


234  perce  nt  age. 

Examples  for  Practice. 

2.  What  is  the  premium  on  a  policy  for  $750,  at  4=%  ? 

Ans.  $30. 

3.  What  premium  must  be  paid  for  $4572.80  insurance, 
at  2J  per  cent.  ?  Ans.  $114.32. 

4.  A  house  and  furniture,  valued  at  $5700,  are  insured  at 
If  per  cent. ;  what  is  the  premium  ?  Ans.  $99.75. 

5.  A  vessel  and  cargo,  valued  at  $28400,  are  insured  at 
3^  per  cent. ;  what  is  the  premium?  Ans.  $994. 

6.  A  woolen  factory  and  contents,  valued  at  $55800,  are 
insured  at  2^  per  cent. ;  if  destroyed  by  fire,  what  would  be 
the  actual  loss  of  the  company  ;  Ans.  $54237.60. 

7.  What  must  be  paid  to  insure  a  steamboat  and  cargo 
from  Pittsburg  to  'New  Orleans,  valued  at  $47500,  at  f  of 
1  per  cent.?  Ans.  $356.25. 

8.  A  gentleman  has  a  house,  insured  for  $8000,  and  the 
furniture  for  $4000,  at  2|  per  cent. ;  what  premium  must 
he  pay  ?  Ans.  $285. 

9.  A  cargo  of  4000  bushels  of  wheat,  worth  $1.20  a  bushel, 
is  insured  at  f  of  1^  per  cent,  on  f  of  its  value  ;  if  the  cargo 
be  lost,  how  much  will  the  owner  of  the  wheat  lose  ? 

Ans.  $1636. 

10.  What  will  it  cost  to  insure  a  factory  valued  at  $21000, 
at  4  per  cent.,  and  the  machinery  valued  at  $15400,  at  |^? 

Ans.  $264.25. 

TAXES. 

283,  A  Tax  is  a  sum  of  money  assessed  on  the  person 
or  property  of  an  individual,  for  public  purposes. 

284.  When  a  tax  is  assessed  on  property,  it  is  apportioned 
at  a  certain  per  cent,  on  the  estimated  value. 

When  assessed  on  the  person,  it  is  apportioned  equally 
among  the  male  citizens  liable  to  assessment,  and  is  called 
a  poll  tax.    Each  person  so  assessed  is  called  a  poll. 

What  is  a  tax?  How  is  a  tax  on  property  apportioned ?  On  the 
person,  how  ? 


TAXES.  235 

285.  Property  is  of  two  kinds — real  estate,  and  per- 
sonal property. 

286,  Real  Estate  consists  of  immovable  property,  such 
as  lands,  houses,  etc. 

28*7.  Personal  Property  consists  of  movable  property, 
such  as  money,  notes,  furniture,  cattle,  tools,  etc. 

288.  An  Inventory  is  a  written  list  of  articles  of  prop- 
erty, with  their  value. 

289.  Before  taxes  are  assessed,  a  complete  inventory  of 
all  the  taxable  property  upon  which  the  tax  is  to  be  levied 
must  be  made.  If  the  assessment  include  a  poll  tax,  then 
a  complete  list  of  taxable  polls  must  also  be  made  out. 

I.  A  tax  of  $3165  is  to  be  assessed  on  a  certain  town ; 
the  valuation  of  the  taxable  property,  as  shown  by  the 
assessment  roll,  is  $600,000,  and  there  are  220  polls  to  be 
assessed  75  cents  each  ;  what  will  be  the  tax  on  a  dollar,  and 
how  much  will  be  A's  tax,  whose  property  is  valued  at  $3750, 
and  who  pays  for  3  polls  ? 

OPERATION. 

$.75  X  220  =  $165,  amount  assessed  on  the  polls. 

$3165  —  $165  =  $3000,  amount  to  be  assessed  on  the  property. 

$3000  ^  $600,000  =  .005,  tax  on  $1. 

$3750  X  .005  =  $18.75,  A's  tax  on  property. 

$.75  X  3  =  $2.25,  A's  tax  on  3  polls. 

$18.75 +  $2.25  =  $21,  amount  of  A's  tax. 

Rule.  I.  Find  the  amount  of  poll  tax,  if  any,  and  sub- 
tract this  sum  from  the  ivhole  amount  of  tax  to  be  assessed. 

II.  Divide  the  sum  to  be  raised  on  property,  by  the  whole 
amount  of  taxable  property,  and  the  quotient  ivill  be  the  per 
cent,,  or  the  tax  on  one  dollar. 

III.  Multiply  each  man^s  taxable  property  by  the  per  cent., 
or  the  tax  on  $1,  and  to  the  product  add  his  poll  tax,  if  any ; 
the  result  ivill  be  the  'whole  amount  of  his  tax. 

What  is  real  estate  ?  Personal  |)ropei-ty  ?  An  inventory  ?  Explain 
the  process  of  levying  a  state  or  other  tax.     Rule. 


236 


P  E  R  C  E  iq^  T  A  G  E . 


Ilaving  found  the  tax  on  $1,  or  the  per  cent.,  which  in  the  preceding  example  we 
find  to  be  5  mills,  or  -^  per  cent.,  the  operation  of  assessing  taxes  maj'  be  greatly 
facilitated  by  finding  the  tax  on  $2,  $3,  etc.,  to  $10,  and  then  on  $20,  $30,  etc.,  to 
$100,  and  arranging  the  numbers  as  in  the  followmg 

Table. 


Prop. 

Tax. 

Prop. 

Tax. 

Prop. 

Tax. 

Prop. 

Tax. 

$1  gives 

$.005 

$10 

$.05 

$100 

$  .50 

$1000 

$  5.00 

2   ** 

.01 

20 

.10 

200 

1.00 

2000 

10. 

3  *' 

.015 

30 

.15 

300 

1.50 

3000 

15. 

4  - 

.03 

•40 

.20 

400 

2.00 

4000 

20. 

5  " 

.025 

50 

.25 

500 

2.50 

5000 

25. 

6   - 

.03 

60 

.30 

600 

8.00 

6000 

30. 

7  " 

.035 

70 

.35 

700 

3.50 

7000 

35. 

8  " 

.04 

80 

.40 

800 

4.00 

8000 

40. 

9   " 

.045 

90 

.45 

900 

4.50 

9000 

45. 

Examples  for  Practice. 

2.  According  to  the  conditions  of  the  last  example,  how 
much  would  he  a  person's  tax  whose  property  was  assessed 
at  $3845,  and  who  paid  for  2  polls? 

Finding  the  amount  from  the  table, 

The  tax  on  $3000 is $15.00 

"     "     ''       800 "  4.00 

"     **     "         40 "  .20 

*'     "     "  5 "  .025 

*'     **     "          2  polls..."  1.50 

Total  tax  is $20,725 

3.  How  much  would  he  Ws  tax,  who  was  assessed  for  1 
poll,  and  on  property  valued  at  $5390  ?  Ans,  $27.70. 

4.  A  tax  of  $9190.50  is  to  be  assessed  on  a  certain  village ; 
the  property  is  valued  at  $1400000,  and  there  are  2981  polls, 
to  be  taxed  50  cents  each  ;  what  is  the  assessment  on  a  dol- 
lar ?  what  is  C's  tax,  his  property  being  assessed  at  $12450, 
and  he  paying  for  2  polls  ? 

Ans.  $.005 J  on  $1  ;  $69.47^,  C's  tax. 

5.  What  is  the  tax  of  a  non-resident,  having  property  in 
the  same  village  valued  at  $5375  ?  Ans.  $29.5625. 

Explain  the  table  and  its  use. 


CUSTOM-HOUSE    BUSINESS.  237 

6.  A  mining  corporation,  consisting  of  30  persons,  are 
taxed  $4342.75  ;  tlieir  property  is  assessed  for  $188000,  and 
each  poll  is  assessed  62J-  cents  ;  what  per  cent,  is  their  tax, 
and  how  much  must  he  pay  whose  share  is  assessed  for 
$2500,  and  who  pays  for  1  poll  ?        A7is.  2-^%  ;  $58,125. 

7.  In  a  certain  county,  containing  25482  taxable  inhab- 
itants, a  tax  of  $103294.60  is  assessed  for  town,  county,  and 
state  purposes ;  a  part  of  this  sum  is  raised  by  a  tax  of  30 
cents  on  each  poll ;  the  entire  valuation  of  property  on  the 
assessment  roll  is  $38260000  ;  what  per  cent,  is  the  tax,  and 
how  much  will  a  person's  tax  be  who  pays  for  3  polls,  and 
whose  property  is  valued  at  $9470  ?    A^is.  to  last,  $24,575. 

8.  The  number  of  polls  in  a  certain  school  district  is  225, 
and  the  taxable  property  $1246093.75  ;  it  is  proposed  to 
build  a  union  school  house  at  an  expense  of  $10000 ;  if  the 
poll  tax  be  $1.25  a  poll,  and  the  cost  of  collecting  be  2J  per 
cent.,  what  will  be  the  tax  on  a  dollar,  and  how  much  will  be 
E's  tax,  who  pays  for  1  poll,  and  has  property  to  the  amount 
of  $11500  ?  Ans.  $.008,  tax  on  $1;  $93.25,  E's  tax. 

9.  In  a  certain  district  the  school  was  supported  by  a  rate- 
bill  ;  the  tea<ihei^s  wages  amounted  to  $200,  the  fuel  and 
other  expenses  to  $75.57  ;  the  public  money  received  was  $98, 
and  the  whole  number  of  day's  attendance  was  3946;  A  sent 
2  pupils  118  days  each  ;  how  much  was  his  rate  bill? 

Ans.  $10.62. 

CUSTOM-HOUSE    BUSINESS. 

290.  Duties,  or  Customs,  are  taxes  levied  on  imported 
goods,  for  the  support  of  government  and  the  protection  of 
home  industry. 

291.  A  Custom-House  is  an  office  established  by  gov- 
ernment for  the  transaction  of  business  relating  to  duties. 

292.  A  Port  of  Entry  is  a  seaport  town  having  a  cus- 
tom-house. 

Define  duties.     A  custom-house. 


238  PEKCENTAGE. 

293.  Tonnage  is  a  tax  levied  upon  a  vessel,  independent 
of  its  cargo,  for  the  privilege  of  coming  into  a  port  of  entry. 

294.  Revenue  is  the  income  to  government  from  duties 
and  tonnage. 

Duties  are  of  two  kinds — ad  valorem  and  specific. 

295.  Ad  Valorem  Duty  is  a  sum  computed  on  tlie  cost 
of  goods  in  the  country  from  which  they  were  imported. 

296.  Specific  Duty  is  a  sum  computed  on  the  weight 
or  measure  of  goods,  without  regard  to  their  cost. 

297.  An  Invoice  is  a  bill  of  goods  imported,  showing 
the  quantity  and  price  of  each  kind. 

298.  By  the  New  Tariff  Act,  approved  March  2,  1857, 
all  duties  taken  at  the  U.  S.  custom-houses,  are  ad  valorem. 

In  collecting  customs,  it  is  the  design  of  government  to 
tax  only  so  much  of  the  merchandise  as  will  be  available  to 
the  importer  in  the  market.  The  goods  are  weighed,  meas- 
ured, gauged,  or  inspected,  in  order  to  ascertain  the  actual 
quantity  and  value  received  in  port ;  and  an  allowance  is 
made  in  every  case  of  waste,  loss,  or  damage. 

299.  Tare  is  an  allowance  of  the  weight  of  the  package 
or  covering  that  contains  the  goods.  It  is  ascertained  by 
actually  weighing  one  or  more  of  the  empty  boxes,  casks, 
or  coverings.  In  common  articles  of  importation,  it  is 
sometimes  computed  at  a  certain  per  cent,  previously  ascer- 
tained by  frequent  trials. 

300.  Leakage  is  an  allowance  on  liquors  imported  in 
casks  or  barrels. 

301.  Breakage  is  an  allowance  on  liquors  imported  in 
bottles. 

Actual  leakage  or  breakage  ie  allowed,  there  being  no  fixed  or  legal  rate. 

302.  Gross  Weight  or  Value  is  the  weight  or  value  of 
the  goods  before  any  allowance  has  been  made. 

303.  Net  Weight  or  Value  is  the  weight  or  value  after 
all  allowances  have  been  deducted. 

Define  Tonnage.  Revenue.  Ad  valorem  duty.  Specific  duty.  An 
invoice.  Tare.  Leakage.  Breakage.  Groas  weight  or  vahio.  Not 
woiglit  or  value. 


CUSTOM-HOUSE    BU  S.I  NESS.  239 

Draft  is  an  allowance  for  the  waste  of  certain  articles,  and  is  made  only  for  statis- 
tical purposes  ;  it  does  not  aflfect  the  amount  of  duty.  The  rates  of  this  allowance  are 
as  follows : 

On         1121b lib. 

Above    112  lb.,  and  not  exceeding    224  lb.,  2  lb. 

2241b.,    "      "         "  336  lb.,  3  lb. 

3361b.,     "      "         "  1120  lb.,  4  lb. 

"      11201b.,    *'      "  *'  2016  lb.,  7  lb. 

"      2016  lb 9  lb. 

What  is  the  duty,  at  24  per  cent.,  on  50  gross  of  London 

ale,  invoiced  at  $1.20  per  dozen,  2J  per  cent,  being  allowed 

for  breakage  ' 

OPKRATiON.  Analysis.    First 

$1.20  X  12  X  50  =  $720,  gross  yalue.         find  the  cost  of  the 

$720  X  .025  =  $18,  breakage.  ^1®'  at  the  invoice 

$720  -  $18  =  $702,  net  value.  T'^^'  ^^""^  ^  ^^?' 

^.^rs^         ^j         ^-ir>r^  Ar^     T    ,  FroHi  tMs  suui  dc- 

1702  X  .34  =  $168.48,  duty.  ^^^^  ,^^  ^1^^^^^ 

for  breakage,  $18,  and  compute  the  duty  on  the  remainder. 

Rule.  Deduct  allowances,  if  necessary,  and  compute  the 
duty,  at  the  given  rate,  on  the  net  value. 

In  the  following  examples,  the  legal  rate  of  duty  will  be  given,  according  to  the 
Tariff  of  1851. 

Examples  for  Practice. 

2.  What  is  the  duty  at  19  per  cent,  on  224  yards  of  plaid 
silk,  invoiced  at  $.95  per  yard  ?  Ans.  $40.43  +  . 

3.  What  is  the  duty  at  24  per  cent,  on  50  barrels  of  sperm 
oil,  each  containing  originally  31|-  gallons,  invoiced  at  $.54 
per  gallon,  allowing  2  per  cent,  for  leakage  ?  Ans.  $200.03  -f. 

4.  What  is  the  duty  at  15  per  cent,  on  175  bags  of  Java 
coffee,  each  containing  115  lbs.,  valued  at  15  cents  per 
pound?  Ans.  $452.81  J. 

5.  John  Jones  imported  from  Havana  25  hhds.  of  W.  I. 
molasses,  which  was  invoiced  at  36  cents  per  gallon  ;  allow- 
ing J  per  cent,  for  leakage,  what  was  the  duty  at  24^  ? 

Ans.  $135,399, 

Define  draft.     Give  analysis.     Rule. 


240 


SIMPLE    I2TTEREST. 


SIMPLE    INTEREST. 

304.  Interest  is  a  sum  paid  for  the  use  of  money. 

305.  Principal  is  the  sum  for  the  use  of  which  interest 
is  paid. 

306.  Rate  per  cent,  per  annum  is  the  sum  per  cent, 
paid  for  the  use  of  $100  annually. 

307.  Amount  is  the  sum  of  the  principal  and  interest. 

308.  Simple  Interest  is  the  sum  paid  for  the  use  of  the 
principal  only,  during  the  whole  time  of  the  loan  or  credit. 

309.  Legal  Interest  is  the  rate  per  cent,  established  by 
law.     It  Taries  in  different  States,  as  follows  : 


state. 

Lejral 
Rate. 

Special 
Agreement. 

state. 

^t. 

Special 

by 

Agreement. 

Alabama 

Arkansas 

8% 
6% 

10% 
7% 

10% 
6% 

10% 
6% 
6% 
8% 
7% 

10% 
6% 
6% 
6% 
7% 
6% 
5% 
6% 
6% 
6% 

Any  rate. 
Any  rate. 
Any  rate. 
Any  rate. 

Any  rate. 

10% 
Any  rate. 

10% 
Any  rate. 

10% 

10% 

10% 

12% 

10% 

8% 

Any  rate. 

Any  rate. 

10% 
12% 

Mississippi 

Missouri 

67o 
6% 

10% 
6% 
6% 
6% 
6% 

10% 

10% 
6% 

10% 
6% 
6% 
7% 
6% 
8% 

10% 

'4 
'4 
% 

10% 

10% 

California 

Montana 

Any  rate. 

'  8  % 

12% 

Any  rate. 

8^ 
12^ 

Connecticut  . . 

Colorado 

Canada 

Dakota 

Delaware 

New  Hampshire. . 

New  Jersey 

New  York 

North  Carolina . . . 
Nebraska 

Dist.  Columbia 

Nevada 

Florida 

Ohio 

Georgia 

Idaho 

Oregon 

Pennsylvania 

Rhode  Island 

South  Carolina 

Tennessee 

Texas 

Illinois 

Indiana 

Iowa 

Any  rate. 
Any  rate. 

10% 
Any  rate. 
Any  rate. 

12% 

10% 
Any  rate. 

Kansas 

Kentucky 

Louisiana 

Utah 

Vermont 

Virginia 

Maryland 

Massachusetts  . 

Michigan 

Minnesota 

West  Virginia. . . . 

Wisconsin 

Washington  Ter. . 
England 

310.  Usury  is  illegal  interest,  or  a  greater  per  cent,  than 
the  legal  rate. 

Case  I. 

311.  To  find  the  interest  on  any  sum,  at  any  rate 
per  cent.,  for  years  and  months. 


PERCEIJTAGE.  241 

In  percentage,  any  per  cent,  of  any  given  number  is  so 
many  hundredths  of  that  number;  but  in  interest,  any  rate 
per  cent,  is  confined  to  1  year,  and  the  per  cent,  to  be  ob- 
tained of  any  given  number  is  greater  than  the  rate  per  cent, 
per  annum  if  the  time  be  more  than  1  year,  and  less  than  the 
rate  per  cent,  per  annum  if  the  time  be  less  than  1  year. 
Thus,  the  interest  on  any  sum,  at  any  rate  per  cent.,  for  3 
years  6  months,  is  3^  times  the  interest  on  the  same  sum 
for  1  year  ;  and  the  interest  for  3  months  is  J  of  the  interest 
for  1  year. 

1.  What  is  the  interest  on  $75. 19  for  3  years  6  months, 
at  6^? 

OPERATION. 
$75.19 

.06  Analysis.    The  interest  on  $75.19  for  1  yr., 

rTTTTT  at  6  per  cent.,  is  .06  of  the  principal,  or  $4.5114, 

and  the  interest  for  3  yr.  6  mo.  is  3y%  =  3^  times 
the  interest  for  1  yr.,  or  $4.5114  x  3|,  which  is 


H 


22557      $15,789+. 
135342 


$15.7899,  Ans, 

Rule.  I.  Multiply  the  principal  hy  the  rate  per  cent,,  and 
the  product  will  he  the  interest  for  1  year, 

II.  Multiply  this  product  hy  the  time  in  years  and  frac- 
tions of  a  year,  and  the  result  will  he  the  required  interest. 
Examples  foe  Pkactice. 

2.  What  is  the  interest  of  $150  for  3  years,  at  4  per  cent? 

Ans.  $18. 

3.  What  is  the  interest  of  $328  for  2  years,  at  1%  ? 

4.  What  is  the  interest  of  $125  for  1  year  6  months,  at 
6^?  Ans.  $11.25. 

5.  What  is  the  interest  of  $200  for  3  years  10  months,  at 
7  per  cent.?  Ans.  $53.66  +  . 

6.  What  is  the  interest  of  $76.50  for  2  years  2  months,  at 

5^?  A71S.  $8.287. 

Explain  the  difference  between    percentage  and  interest      Give 
analysis.     Rule. 
i:.r.  11 


242  SIMPLE    INTEREST. 

7.  What  is  the  interest  of  11276.25  for  11  months,  at  7 
percent?  Ans,  $81.89  +  . 

8.  What  is  the  interest  of  $2569.75  for  4  years  6  months, 
at  6  per  cent.  ? 

9.  What  is  the  interest  of  $1500.60  for  2  years  4  months, 
at  6i^?  Ans.  $218.8375. 

10.  What  is  the  amount  of  $26.84  for  2  years  6  months, 
at  5  per  cent.  ?  Ans.  $30,195. 

11.  What  is  the  amount  of  $450  for  5  years,  at  7  per  cent.  ? 

12.  What  is  the  interest  of  $4562.09  for  3  years  3  months, 
at  3^?  Ans.  $444.80  +  . 

13.  What  is  the  amount  of  $3050  for  4  years  8  mouths,  at 
5{  per  cent.  ?  Ans.  $3797.25+. 

14.  What  is  the  interest  of  $5000  for  9  months,  at  8  per 
cent.?  Ans.  $e'500. 

15.  If  a  person  borrow  $375  at  7  per  cent.,  how  much  will 
be  due  the  lender  at  the  end  of  2  yr.  6  mo.  ? 

16.  What  is  the  interest  paid  on  a  loan  of  $1374.74,  at  6 
per  cent.,  made  January  1,  1856,  and  called  in  January  1, 
1860?  Ans.  $329,937. 

17.  If  a  note  of  $605.70  given  May  20, 1858,  on  interest  at 
8  per  cent,  be  taken  up  May  20, 1861,  what  amount  will  then 
be  due  if  no  interest  has  been  paid  ?  Ans,  $751,068. 

Case  IL 
312.   To  find  the  interest  on  any  sum,  for  any 
time,  at  any  rate  per  cent. 
The  analysis  of  our  rule  is  based  upon  the  following 
Obvious  Relations  between  Time  and  Interest 

I.  The  interest  on  any  sum,  for  1  year,  at  1  per  cent.,  is 
.01  of  that  sum,  and  is  equal  to  the  principal  with  the 
separatrix  removed  two  places  to  the  left. 

II.  A  month  being  ^  of  a  year,  -^^g^  of  the  interest  on  any 
sum  for  1  year  is  the  interest  for  1  month. 

What  is  Case  II  ?  Give  the  first  relation  between  time  and  interest. 
Second. 


PERCENTAGE.  243 

III.  The  interest  on  any  sum  for  3  days  is  -^^  =  yig-  =  .1 
of  the  interest  for  1  month,  and  any  number  of  days  may 
readily  be  reduced  to  tenths  of  a  month  by  dividing  by  3. 

IV.  The  interest  on  any  sum,  for  1  month,  multiplied  by 
any  given  time  expressed  in  months  and  tenths  of  a  month, 
will  produce  the  required  interest. 

1.  What  is  the  interest  on  1724.68  for  2  yr.  5  mo.  19  da., 
at  7^? 

OPERATION.  Analysis.    We  remove 

2  yr.  5  mo.  19  da.  =  29.6-J  mo.  the  separatrix  in  tlie  given 

12  ^  $7  2468  principal   two  places  to  tlie 

'- left,  and  have   $7.2468,  the 

^.o0o9  interest  on  the  given  sum  for 

29.6^  1  year  at  1  per  cent.  (313, 

OA-j  o  I.).     Dividing  this  by  12,  we 

have  $.6039,  the  interest  for 

^"^^^  1  month,  at  1  per  cent. 

S^^^l  (II.)  Multiplying  this  quo- 

12078  tient  by  29.6^,  the  time  ex- 

^17  89557  pressed  in  months  and  deci- 

-,  mals  of  a  month,  (III.,  IV.,) 

*  we  have  $17.89557,  the  in- 


$125.26899,  Ans.  terest  on  the  given  sum  for 

the  given  time,  at  1  per  cent. 

(IV.).     And  multiplying  this  product  by  7  (7  times  1  per  cent.),  we 

have  $125,268,  the  interest  on  the  given  principal,  for  the  given  time, 

at  the  given  rate  per  cent. 

KuLE.  I.  Remove  the  separatrix  in  the  given  principal 
tioo  places  to  the  left;  the  result  will  he  the  interest  for  1 
year,  at  1  per  cent, 

II.  Divide  this  interest  ly  12 ;  the  result  will  he  the  interest 
for  1  month,  at  1  per  cent. 

III.  Multiply  this  interest  hy  the  given  time  expressed  in 
months  and  tenths  of  a  month  ;  the  result  ivill  he  the  interest 
for  the  given  time,  at  1  per  cent, 

IV.  Multiply  this  interest  hy  the  given  rate ;  the  product 
loill  he  the  interest  required. 

Give  the  third.     Fourth.     Give  analysis.     Rule. 


244  SIMPLE    INTEREST. 

Contractions.  After  removing  the  separatrix  in  the  principal,  two 
places  to  the  left,  the  result  may  be  regarded  either  as  the  interest  on 
the  given  principal  for  12  months  at  1  per  cent.,  or  for  1  month  at  12 
per  cent.  If  we  regard  it  as  for  1  month  at  12  per  cent. ,  and  if  the 
given  rate  be  an  aliquot  part  of  12  per  cent.,  the  interest  on  the  given 
principal  for  1  month  may  readily  be  found  by  taking  such  an  aliquot 
part  of  the  interest  for  1  mouth  as  the  given  rate  is  part  of  12  per 
cent.     Thus, 

To  find  the  interest  for  1  month  at  6  per  cent.,  remove  the  separa- 
trix two  places  to  the  left,  and  divide  by  2. 

To  find  it  at  3  per  cent.,  proceed  as  before,  and  divide  by  4 ;  at  4  per 
cent.,  divide  by  3 ;  at  2  per  cent.,  divide  by  6,  etc. 

Six  Pee  Cent.  Method. 

313.  By  referring  to  309  it  will  be  seen  that  the  legal 

rate  of  interest  in  21  States  is  6  per  cent.  This  is  a  sufficient 
reason  for  introducing  the  following  brief  method  into  this 
work  : 

Analysis.    At  6  per  cent,  per  annum  the  interest  on  $1 
For  12  months is  $.06. 

"      2  months  (3%  =  i  of  12  mo.) "     .01. 

'•      1  month,  or  30  days  (yV  of  12  mo.). .  "     .00^  =  $.005  (^  of  $.06). 

"      6  days  (^  of  30  days) "    .001. 

"      1    "     (^  of  6  da.  =  ^V  of  30  days) . .  "     .000^. 
Hence  we  conclude  that, 

1st.  The  interest  on  $1  is  $.005  per  month,  or  $.01  for 
every  2  months ; 

2d.  The  interest  on  $1  is  $.000|  per  day,  or  $.001  for 
every  6  days. 

From  these  principles  we  deduce  the 

Rule.  I.  To  find  the  rate  : — Call  every  year  $.06,  every 
2  months  $.01,  every  6  days  $.001,  and  every  less  number  of 
days  sixths  of  1  mill. 

II.  To  find  the  interest : — Multiply  the  principal  hy  the 
rate. 

1.  To  find  the  interest  at  any  other  rate  per  cent,  by  this  method,  first  find  it  at 
6  per  cent.,  and  then  increase  or  diminish  the  result  by  as  many  times  itself  as  the 

What  contractions  are  given  ?  Give  analysis  of  the  6  per  cent 
method.     Rule.     Its  application  to  any  other  rate  per  cent. 


SIMPLE     INTEREST.  245 

^ven  rale  is  greater  or  less  than  6  per  cent.    Thus,  for  7  per  cent,  add  |,  for  4  per 
cent,  subtract  i,  etc. 

2.  The  interest  of  $10  for  6  days,  or  of  $1  for  60  days,  is  $.01.  Therefore,  if  the 
principal  be  less  than  $10  and  the  time  less  than  6  days,  or  the  principal  less  than 
$1  and  the  time  less  than  60  days,  the  interest  will  be  less  than  $.01,  and  may  be 
disregarded. 

3.  Since  the  inte^st  of  $1  for  60  days,  is  $.01,  the  interest  of  $1  for  any  number 
of  days  is  as  many  cents  as  60  is  contained  times  in  the  number  of  days.  Therefore, 
if  any  principal  be  multiplied  by  the  number  of  days  in  any  given  number  of  months 
and  days,  and  the  product  divided  by  60,  the  result  will  be  the  interest  in  cents. 
That  is.  Multiply  the  principal  by  the  number  of  days,  divide  the  product  by  60,  and 
point  off  two  decimal  places  in  the  quotient.  T/ie  result  wHl  be  the  interest  in  the 
8am£  denomination  as  the  principal. 

Examples  for  Practice. 

2.  What  is  the  interest  of  $100  for  7  years  7  months,  at 
6  per  cent.  ?  Ans.  145.50. 

3.  What  is  the  amount  of  147.50  for  4  years  1  month,  at 
9  per  cent.  ?  Ans.  164.956. 

4.  What  is  the  amount  of  $2000  for  3  months,  at  7  per 
cent?  A71S.  $2035. 

5.  What  is  the  interest  of  $250  for  1  year  10  months  and 
15  days,  at  6  per  cent.  ?  Ans.  $28. 12 J. 

6.  What  is  the  interest  of  $36.75  for  2  years  4  months  and 
12  days,  at  7%  ?  Ans.  $6,088. 

7.  What  is  the  amount  of  $84  for  5  years  5  months  and 
9  days,  at  5  per  cent.  ? 

8.  What  is  the  interest  of  $51.10  for  10  months  and  3 
days,  at  4^  ? 

9.  What  is  the  interest  of  $175.40  for  15  months  and  8 
days,  at  10  per  cent?  Ans.  $22.31  +  . 

10.  What  is  the  amount  of  $1500  for  6  months  and  24 
days,  at  7^^?  Ans.  $1563.75. 

11.  What  is  the  amount  of  $84.25  for  1  year  5  months 
and  10  days,  at  6J  per  cent.  ? 

12.  What  is  the  interest  of  $25  for  3  years  6  months  and 
20  days,  at  6  per  cent  ?  Ans.  $5.33|. 

13.  What  is  the  interest  of  $112.50  for  3  months  and  1 
day,  at  9J^?  Ans.  $2.70+. 

What  contractions  are  given  ? 


246  PERCENTAGE. 

14.  What  is  the  interest  of  $408  for  20  days,  at  6  per 
cent.?    *  Ans.  $1.36. 

15.  What  is  the  interest  of  $500  for  22  days,  at  7  per  cent.  ? 

16.  What  is  the  amount  of  $4500  for  10  days,  at  10  per 
cent.?  Ans.  $4512.50. 

17.  What  is  the  amount  of  $1000  for  1  month  5  days,  at 
6|  per  cent.  ?  Ans.  $1006. 56 J. 

18.  Find  the  interest  of  $973.68  for  7  months  9  days,  at 

19.  If  I  borrow  $275  at  7  per  cent.,  how  much  will  I  owe 
at  the  end  of  4  months  25  days  ? 

20.  A  person  bought  a  piece  of  property  for  $2870,  and 
agreed  to  pay  for  it  in  1  year  and  6  months,  with  Q^  per  cent, 
interest ;  what  amount  did  he  pay?  Ans,  $3149.825. 

21.  In  settling  with  a  merchant,  I  gave  my  note  for  $97.75, 
due  in  11  months,  at  5  per  cent. ;  what  must  be  paid  when 
the  note  falls  due  ?  Ans.  $102. 23  + . 

22.  How  much  is  the  interest  on  a  note  of  $384.50  in  2 
years  8  months  and  4  days,  at  S%  ? 

23.  What  is  the  interest  of  $97.86  from  May  17,  1850,  to 
December  19,  1857,  at  7  per  cent.?  A^is.  $51.98  + . 

24.  Find  the  interest  of  $35.61,  from  Nov.  11,  1857,  to 
Dec.  15,  1859,  at  6  per  cent.  A^is.  $4,474. 

25.  Required  the  interest  of  $50  from  Sept.  4,  1848,  to 
Jan.  1,  1860,  at  3^%. 

26.  Required  the  amount  of  $387.20,  from  Jan.  1,  to  Oct. 
20,  1859,  at  7  per  cent.  Ans.  $408,957. 

27.  A  man,  owning  a  furnace,  sold  it  for  $6000  ;  the  terms 
were,  $2000  m  cash  on  delivery,  $3000  in  9  months,  and  the 
remainder  in  1  year  6  months,  with  7  per  cent,  interest ; 
what  was  the  whole  amount  paid  ?  *  Ans.  $6262.50. 

28.  Wm.  Gallup  bought  bills  of  dry  goods  of  Geo.  Bliss 
&  Co.,  of  New  York,  as  follows,  viz.  :  Jan.  10,  1858,  $350; 
April  15, 1858,  $150  ;  and  Sept.  20, 1858,  $550.50  ;  he  bought 
on  time,  paying  legal  interest ;  what  was  the  whole  amount 
of  his  indebtedness  Jan.  1,  1859?  Ans.  $1092.66  +  . 


PARTIAL    PAYMENTS.  247 


PARTIAL  PAYMENTS  OR  INDORSEMENTS. 

314.  A  Partial  Payment  is  payment  in  part  of  a  note, 
bond,  or  other  obligation ;  when  the  amount  of  a  payment 
is  written  on  the  back  of  the  obligation,  it  becomes  a  re- 
ceipt, and  is  called  an  Indorsement, 

^^QQQ-  Sprikgfield,  Mass.,  Jan.  4,  1857. 

1.  For  value  received  I  promise  to  pay  James  Parish,  or 
order,  two  thousand  dollars,  one  year  after  date,  with  in- 
terest. George  Joi^es. 

On  this  note  were  indorsed  the  following  payments: 

Feb.  19,   1858 $400 

June  29,  1859 $1000 

Nov.  14,  1859 $520 

What  remained  due  Dec.  24,  1860  ? 

OPERATION. 

Principal  on  interest  from  Jan.  4,  1857 $2000 

Interest  to  Feb.  19,  1858,  1  yr,  1  mo.15  da 135 

Amount $2135 

Payment  Feb.  19,  1858 400 

Eemainder  for  a  new  principal $1735 

Interest  from  Feb.  19,  1858,  to  June  29,  1859,  1  yr.  4  mo. 

10  da 141.69 

Amount $1876.69 

Payment  June  29, 1859 1000 

Eemainder  for  a  new  principal $876.69 

Interest  from  June  29,  1859,  to  Nov.  14, 1859, 4  mo.  l5  da.       19.725 

Amount $896,415 

Payment  Nov.  14,  1859 520 

Remainder  for  a  new  principal $376,415 

Interest  from  Nov.  14, 1859,  to  Dec.  24,  1860, 1  yr.  1  mo. 

10  da 25.09 

Remains  due  Dec.  24, 1860 $401,505  + 

What  is  meant  by  partial  payment  ?    By  an  indorsement  ? 


248  PERCENTAGE. 


^^^^5.50.  jSTew  York,  May  1,  1855. 

2.  For  value  received,  we  jointly  and  severally  promise  to 
pay  Mason  &  Bro.,  or  order,  four  hundred  seventy-five  dol- 
lars fifty  cents,  nine  months  after  date,  with  interest. 

Jokes,  Smith  &  Co. 

The  following  indorsements  were  made  on  this  note  : 

Dec.  25,  1855,  received $50 

July  10,  1856,        ''     15.75 

Sept.    1,  1857,        ''      25.50 

June  14,  1858,        ''      104 

How  much  was  due  April  15,  1859  ? 


OPERATION. 

Principal  on  interest  from  May  1, 1855 $475.50 

Interest  to  Dec.  25,  1855,  7  mo.  24  da 21.63 

Amount $497.13 

Payment  Dec.  25, 1855 50 

Remainder  for  a  new  pri.ncipal $447.13 

Interest  from  Dec.  25,  1855,  to  Jime  14,  1858,  2  yr.  5  mo. 

19  da 77.29 

Amount .- $524.42 

Payment  July  10, 1856,  less  than  interest  then  due, )  $15.75 

Payment  Sept.  1,  1857 S    25.50 

Tlieir  sum  less  than  interest  then  due $41.25 

Payment  June  14, 1858 104 

Their  sum  exceeds  the  interest  then  due $145.25 

Remainder  for  a  new  principal $379.17 

Interest  from  June  14,  1858,  to  April  15. 1859, 10  mo.  1  da..     22.19 
Balance  due  April  15, 1859 $401.36  + 

These  examples  have  been  wrought  according  to  the 
method  prescribed  by  the  Supreme  Court  of  the  U.  S.,  and 
are  sufificicnt  to  illustrate  the  following 


PARTIAL    PAYMEI^Ta.  249 

TJkited  States  Eule> 

I.  Find  the  amount  of  the  given  principal  to  the  time  of 
the  first  payment,  and  if  this  payment  exceed  the  interest 
then  due  subtract  it  from  the  amount  obtained,  and  treat 
the  remainder  as  a  new  principal. 

II.  But  if  the  interest  be  greater  than  any  payment,  cast 
the  interest  on  the  same  principal  to  a  time  when  the  sum  of 
the  payments  shall  eqtml  or  exceed  the  interest  due  ;  subtract- 
ing the  Slim  of  the  payments  from  the  amount  of  the  princi- 
pal, the  remainder  loill  form  a  new  principal,  on  ivhich  inter- 
est is  to  be  computed  as  before. 


^514.96.  S^i^  Feakcisco,  June  20,  1858. 

3.  Three  years  after  date  we  promise  to  pay  Ross  &  TVade, 
or  order,  five  hundred  fourteen  and  -f^  dollars,  for  value 
received,  with  10  per  cent,  interest.  Wilder  &  Bro. 

On  this  note  were  indorsed  the  following  payments  :  Nov. 
12,  1858,  $105.50  ;  March  20,  1860,  $200  ;  July  10,  1860, 
$75.60.  How  much  remains  due  on  the  note  at  the  time  of 
its  maturity  ?  Ans.  $242.12  + , 

^'^QQQ-  Charleston-,  May  7,  1859. 

4.  For  value  received,  I  promise  to  pay  George  Babcocl 
three  thousand  dollars,  on  demand,  with  7  per  cent,  interest 

JoHK  May. 

On  this  note  were  endorsed  the  following  payments  : 

Sept.  10,  1859,  received $25 

Jan.     1,  1860,        ''      500 

Oct.    25,  1860,        ''      75 

April   4,  1861,        ''      1500 

How  much  was  due  Feb.  20,  1862  ?     Ans.  $1344.35  -f. 

Give  the  United  States  Court  rule  for  computing  interest  where  par- 
tial payments  have  been  made. 

11* 

^  / 


250       .  PERCENTAGE. 


$912^^,  ]sfEw  Orleans,  Aug.  3,  1850. 

5.  One  year  after  date  I  promise  to  pay  George  Bailey,  or 
order,  nine  hundred  twelve  ^^  dollars,  with  5  per  cent,  in- 
terest, for  value  received.  James  Powell. 

The  note  was  not  paid  when  due,  but  was  settled  Sept. 
15,  1853,  one  payment  of  $250  having  been  made  Jan.  1, 
1852,  and  another  of  $316.75,  May  4,  1853.  How  much 
was  due  at  the  time  of  settlement  ?  Ans.  $467.53+. 

^184.56.  Cincinnati,  April  2,  1860. 

6.  Four  months  after  date  I  promise  to  pay  J.  Ernst  & 
Co.  one  hundred  eighty-four  dollars  fifty-six  cents,  for  value 
received.  S.  Anderson. 

The  note  was  settled  Aug.  26,  1862,  one  payment  of  $50 
having  been  made  May  6,  1861.  How  much  was  due,  legal 
interest  being  6%  ?  Ans,  $154,188. 

a  note  is  on  interest  after  it  becomes  due,  if  it  contain  no  mention  of  interest. 

7.  Mr.  B.  gave  a  mortgage  on  his  farm  for  $6000,  dated 
Oct.  1,  1851,  to  be  paid  in  6  years,  with  8  per  cent,  inter- 
est. Three  months  from  date  he  paid  $500  ;  Sept.  10, 1852, 
$1126  ;  March  31, 1854,  $2000  ;  and  Aug.  10, 1854,  $876.50. 
How  much  was  due  at  the  expiration  of  the  time  ? 

Ans.  $3284.84+. 
315.  The  United  States  rule  for  partial  payments  has 
been  adopted  by  nearly  all  the  States  of  the  Union ;  the 
only  prominent  exceptions  are  Connecticut,  Vermont,  and 
New  Hampshire. 

Connecticut  Rule. 
I.  Payments  made  one  year  or  more  from  the  time  the  in 
terest  commenced,  or  from  another  payment,  a7id  paymerits 
less  than  the  interest  due.  are  treated  according  to  the  United 
States  rule. 

(Nearly  obsolete.    The  United  States  rule  is  in  general  use.) 

Give  Connocticut  rule  for  partial  payments. 


PARTIAL    PAYMEN^TS.  251 

II.  Payments  exceeding  the  interest  due,  and  made  toithin 
one  year  from  the  time  interest  commenced,  or  from  a  former 
fayment,  shall  draw  interest  for  the  balance  of  the  year,  pro- 
vided the  interval  does  not  extend  beyond  the  settlement,  and 
the  amount  must  be  subtracted  from  the  amount  of  the  prin- 
cipal for  one  year  ;  the  remainder  will  be  the  new  principal. 

III.  If  the  year  extend  beyond  the  settlement,  then  find  the 
amount  of  the  payment  to  the  day  of  settlement,  and  subtract 
it  from  the  amount  of  the  principal  to  that  day ;  the  remain- 
der luill  be  the  sum  due. 


$460. 


Woodstock,  Ct.,  Jan.  1,  1858. 


1.  For  value  received,  I  promise  to  pay  Henry  Bowen,  or 
order,  four  hundred  sixty  dollars,  on  demand,  with  interest. 

James  Makshall. 

On  this  note  are  indorsed  the  following  payments  :  April 
16,  1858,  $148;  March  11,  1860,  $75;  Sept.  21,  1860,  $56. 
How  much  was  duo  Dec.  11,  1860?  Ans.  $238.15  +  . 

316.  A  note  containing  a  promise  to  pay  interest  an- 
nually is  not  considered  in  law  a  contract  for  anything 
more  than  simple  interest  on  the  principal.  For  partial 
payments  on  such  notes,  the  following  is  the 

Vermont  Eule. 

I.  Find  the  amount  of  the  principal  from  the  time  interest 
commenced  to  the  time  of  settlement. 

II.  Find  the  amount  of  each  payment  from  the  time  it  was 
made  to  the  time  of  settlement. 

III.  Subtract  the  sum  of  the  amounts  of  the  payments  from 
the  amount  of  the  principal,  and  the  remainder  loill  be  the 
sum  due. 


$600. 


EuTLAND,  April  11,  1856. 
1.  For  value  received,  I  promise  to  pay  Amos  Getting,  or 
order,  six  hundred  dollars  on  demand,  with  interest  an- 
nually. John"  Brown. 

Give  the  Vermont  rule  for  partial  payments. 


252  PEECENTAOE. 

On  this  note  were  indorsed  the  following  payments  :  Aug. 
10,  1856,  $156  ;  Feb.  12,  1857,  $200  ;  June  1,  1858,  $185. 
What  was  due  Jan.  1,  1859  ?  A7is,  $105.50  +  . 

317.  In  New  Hampshire  interest  is  allowed  on  the  an- 
nual interest  if  not  paid  when  due,  in  the  nature  of  dam- 
ages for  its  detention ;  and  if  payments  are  made  before  one 
year's  interest  has  occurred,  interest  must  be  allowed  on 
such  payments  for  the  balance  of  the  year. 

New  Hampshiee  Eule. 

I.  Find  the  amount  of  the  principal  for  one  year,  and  de- 
duct from  it  the  amount  of  each  payment  of  that  year,  from 
the  time  it  was  made  up  to  the  end  of  the  year ;  the  remain- 
der will  he  a  neio  principal,  with  ivhich  proceed  as  before, 

II.  If  the  settlement  occur  less  than  a  year  from  the  last 
annual  term  of  interest,  make  the  last  term  of  interest  a  part 
of  a  year,  accordingly. 


$575.  Keei^e,  N.  H.,  Aug.  4,  1858. 

1.  For  value  received,  I  promise  to  pay  G-eorge  Cooper,  or 
order,  five  hundred  seventy-five  dollars,  on  demand,  with 
interest  annually.  David  Geeekmak. 

On  this  note  were  indorsed  the  following  payments  :  Nov. 
4,  1858,  $64 ;  Dec.  13,  1859,  $48 ;  March  16,  1860,  $248  ; 
Sept.  28,  1860,  $60.  What  was  due  on  the  note  Nov.  4, 
1860  ?  •  Ans.  $215.33. 

318.  When  no  payment  whatever  is  made,  upon  a  note 
promising  annual  interest,  till  the  day  of  settlement,  in 
New  Hampshire  the  following  is  the 
Court  Rule. 

Compute  separately  the  interest  on  the  principal  from  the 
time  the  note  is  given  to  the  time  of  settlement,  and  the  inter- 
est on  each  yearns  interest  from  the  time  it  should  be  paid  to 
the  time  of  settlement.  TJie  sum  of  the  hiterests  thus  obtained, 
added  to  the  principal,  will  be  the  sum  due. 

The  New  Hampshire  rule.    The  New  Hampshire  court  rule. 


PARTIAL     PAYMENTS.  253 


^500.  KEEi^E,  N.  H.,  Feb.  2,  1855. 

1.  Three  years  after  date,  I  promise  to  pay  James  Clark, 
or  order,  five  hundred  dollars,  for  value  received,  with  in- 
terest annually  till  paid.  JoHK  S.  Briggs. 

What  is  due  on  the  above  note,  Aug.  2, 1859  ?  Ans.  $649.40. 

Problems  ix  Ii^terest. 

319,  In  examples  of  interest  there  are  five  parts  involved, 
the  Principal,  the  Rate,  the  Time,  the  Interest,  and  the 
Amount. 

Case  I. 

320.  The  time,  rate  per  cent.,  and  interest  being 
given,  to  find  the  principal. 

1.  What  principal  in  2  years,  at  8  per  cent.,  will  gain 
131.80  interest  ? 

OPERATION.  Analysis.    Since  $1,  in 

1.12,  interest  of  $1  in  2  year,  at  6^.  2  years,  at  6  per  cent.,  will 

$31.80  -r-  .12  =  $265,  Ans.  ^^  ^-^^  interest,  the  prin- 

cipal that  will  gain  $31.80, 
at  the  same  rate  and  time,  must  be  as  many  dollars  as  $.12  is  contained 
times  in  $31.80  ;  dividing,  we  obtain  $265,  the  required  principal. 

EuLE.  Divide  the  given  interest  hy  tlie  interest  of  %1  for 
the  given  time  and  rate,  and  the  quotient  will  be  the  principal. 

Examples  for  Practice. 

2.  What  principal,  at  6  per  cent.,  will  gain  $28.1 2^  in  6 
years  3  months  ?  Ans,  $75. 

3.  What  snm,  put  at  interest  for  4  months  18  days,  at  4 
per  cent.,  will  gain  $9.20?  Ans.  $600. 

4.  What  sum  of  money,  invested  at  7  per  cent.,  will  pay 
me  an  annual  income  of  $1260  ?  Ans.  $18000. 

5.  What  sum  must  be  invested  in  real  estate,  yielding  10 
per  cent,  profit  in  rents,  to  produce  an  income  of  $3370  ? 

Ans.  $33700. 


How  many  parts  are  considered  in  examples  in  interest 
they  ?     What  is  Case  I  ?     Give  analysis.     Rule. 


?    What  are 


254  percektage. 

Case  II. 

821.  The  time,  rate  per  cent.,  and  amount  being 
given,  to  find  the  principal. 

1.  What  principal  in  2  years  6  months,  at  7  per  cent., 
will  amount  to  188.125? 

OPEEATION.  AlTALYSIS. 

$1,175  Amt.  of  $1   in  2  years  6  months,  at  7?S.  Since  $1,  in 

$88,125-^1.175  =  175,  Ans,  2   years  6 

months,  at  7 
per  cent.,  will  amount  to  $1,175,  the  principal  that  will  amount  to 
$88,125,  at  the  same  rate  and  time,  must  be  as  many  dollars  as  $1,175 
is  contained  times  in  $88,125  ;  dividing,  we  obtain  $75,  the  required 
principal. 

Rule.   Divide  the  given  amount  hy  the  amount  of  $1  for 
the  given  time  and  rate,  and  the  quotient  will  he  the  principal 


Examples  for  Practice. 

2.  What  principal,  at  6  per  cent.,  wiU  amount  to  $655.20 
in  8  months  ?  Ans.  $630. 

3.  What  principal,  at  5  per  cent.,  wiU  amount  to  $106,855 
in  5  years  5  months  and  9  days?  Ans,  $84. 

4.  What  sum,  put  at  interest,  at  5 J  per  cent.,  for  8  years 
5  months,  will  amount  to  $1897.545  ?     Ans.  $1297.09  +  . 

5.  What  sum,  at  7  per  cent.,  will  amount  to  $221,075  in 
3  years  4  months  ?  Ans.  $179.25. 

6.  What  is  the  interest  of  that  sum,  for  11  years  8  days, 
at  10|-  per  cent.,  which  will  at  the  given  rate  and  time 
amount  to  $857.54?  ^/^5.  $460.04. 

Case  III. 

822.  The  principal,  time,  and  interest  being 
given,  to  find  the  rate  per  cent. 

1.  I  lent  $450  for  3  years,  and  received  for  interest  $67.50 ; 
what  was  the  rate  per  cent.  ? 

Give  Case  II.     Analysis.    Rule.    Case  III. 


PROBLEMS    IN     INTEREST.  255 

OPERATION.  Analysis.     Since  at 

$  4.50  1  per  cent.  $450,  in  3 

g  years,  will  gain  $13.50 

interest,   the  rate  per 

$13.50,  int.  of  $450  for  3  years,  at  \%,  ^^^^  ^^  ^^^^^  ^1^^  ^^^^ 

$67.50  -T-  13.50  =  h%,  A?IS.  principal,  in  tlie  same 

time,  will  gain  $67.50, 
mnst  be  equal  to  the  number  of  times  $13.50  is  contained  in  $67.50; 
dividing,  we  obtain  5,  the  required  rate  per  cent. 

EuLE.  Divide  the  given  interest  ly  the  iiiterest  on  the  prin- 
cipal for  the  given  time  at  1  per  cent.,  and  the  quotient  will 
le  the  rate  per  cent  required. 

Examples  for  Practice. 

2.  If  I  pay  $45  interest  for  the  use  of  $500  for  3  years, 
what  is  the  rate  per  cent.  ?  Ans.  3. 

3.  The  interest  of  $180  for  1  year  2  months  6  days  is 
$12.78?  what  is  the  rate  %?  Ans.  6. 

4.  A  man  invests  $2000  in  bank  stock,  and  receives  a 
semi-annual  dividend  of  $75  ;  what  is  the  rate  per  cent.  ? 

5.  At  what  per  cent,  must  $1000  be  loaned  for  3  years  3 
months  and  29  days,  to  gain  $183.18  ?  Ans.  h^. 

6.  A  man  builds  a  block  of  stores  at  a  cost  of  $21640,  and 
receives  for  them  an  annual  rent  of  $2596.80  ;  what  per  cent, 
does  he  receive  on  the  investment  ?  Ans,  12. 

Case  IV. 
323.  Principal,  interest,  and  rate  per  cent,  being 
given,  to  find  the  time. 

1.  In  what  time  will  $360  gain  $86.40  interest,  at  Q%  ? 

OPERATION.  Analysis.     Since  in 

$  360  1  year  $360,  at  6  per 

QQ  cent.,  will  gain  $21.60, 

the  number  of  years  in 

$21.60,  interest  of  $360  in  1  year,  at  6^.  ^liich  the  same  princi- 

$86.40  ~-  21.60  =  4  years,  Ans.  pal,  at  the  same  rate, 

■will  gain  $86.40,  will  be 

Analysis.     Rulo.     Case  TV.     Analysis. 


356  PERCENTAGE. 

as  many  as  $21.60  is  contained  times  in  $86.40 ;  dividing,  we  obtain 
4  years,  the  required  time. 

EuLE.  Divide  the  given  interest  hy  the  interest  on  the 
principal  for  1  year^  and  the  quotient  will  he  the  time  re- 
quired in  years  and  decimals. 

The  decimal  part  of  the  quotient,  if  any,  may  be  reduced  to  months  and  days  (by 
215). 

Examples  for  Practice. 

2.  The  interest  of  1325  at  6  per  cent,  is  $58.50  ;  what  is 
the  time  ?  Ans.  Z  years. 

3.  B  loaned  $1600  at  6  per  cent,  until  it  amounted  to 
$2000  ;  what  was  the  time  ?  A^is,  4  years  2  months. 

4.  How  long  must  $204  be  on  interest  at  7  per  cent. ,  to 
amount  to  $217.09  ?  Ans.  11  months. 

5.  Engaging  in  business,  I  borrowed  $750  of  a  friend  at 
6  .per  cent.,  and  kept  it  until  it  amounted  to  $942  ;  how  long 
did  I  retain  it  ?  Ans.  4c  years  3  months  6  days. 

6.  How  long  will  it  take  $200  to  double  itself  at  6  per 
cent,  simple  interest?  Ans.  16  years  8  months. 

7.  In  what  time  will  $675  double  itself  at  5^? 

The  time  in  yearB  in  which  any  sum  will  double  itself  may  be  found  by  dividing 
100  by  the  rate  per  cent. 


COMPOUND    INTEREST. 

324.  Compound  Interest  is  interest  on  both  principal 
and  interest,  when  the  interest  is  not  paid  when  due. 

The  simple  interest  may  be  added  to  the  principal  annually,  semi-annually,  or 
quarterly,  as  the  parties  may  agree ;  but  the  taking  of  compound  interest  is  not 
Ugod. 

1.  What  is  the  compound  interest  of  $200,  for  3  years,  at 
6  per  cent.  ? 

Rule.  In  what  time  will  any  8um  double  itself  at  interest  ?  What 
is  compound  interest  ? 


COMPOUND    INTEEEST.  257 

OPERATION. 

$200  Principal  for  1st  year. 

$200  X  .06  = 12  Interest  for  1st  year. 

$212  Principal  for  2d  year.  m 

$212  X  .06  =      12.72  Interest  for  2d  year. 

$224.72    Principal  for  3d  year. 
$224.72  X  .06  =      13.483  Interest  for  3d  year. 

$238,203  Amount  for  3  years. 
200.000  Given  principal. 

$38,203  Compound  interest. 

Rule.  I.  Find  the  amount  of  the  given  principal  at  the 
given  rate  for  one  year,  and  mahe  it  the  principal  for  the 
second  year, 

II.  Find  the  amount  of  this  new  principal,  and  make  it  the 
principal  for  the  third  year,  and  so  continue  to  do  for  the 
given  number  of  years, 

III.  Subtract  the  given  principal  from  the  last  amount,  and 
the  remainder  will  be  the  compound  interest, 

1.  When  the  interest  is  payable  semi-annually  or  quarterly,  find  the  amount  of  the 
given  principal  for  the  first  intei-val,  and  make  it  the  principal  for  the  second  inter- 
val, proceeding  in  all  respects  as  when  the  interest  is  payable  yearly. 

2.  When  the  time  contains  years,  months,  and  days,  find  the  amount  for  the 
years,  upon  which  compute  the  interest  for  the  months  and  days,  and  add  it  to  the 
last  amount,  before  subtracting. 

Examples  for  Practice. 

2.  What  is  the  compound  interest  of  $500  for  2  years  at 
7  per  cent.?  Aiis.  $72.45. 

3.  What  is  the  amount  of  $312  for  3  years,  at  6  per  cent, 
compound  interest  ?  Ans.  $371.59  +  . 

4.  What  is  the  compound  interest  of  1250  for  2  years, 
payable  semi-annually,  at  6  per  cent.?        Ans.  131.37 +  . 

5.  What  will  $450  amount  to  in  1  year,  at  7  per  cent,  com- 
pound interest,  payable  quarterly  ?  Ans,  $482.33. 

6.  What  is  the  compound  interest  of  $236  for  4  years  7 
months  and  6  days,  at  6^  ?  A7is.  $72.66  +  . 

Explain  operation.     Give  rule. 


258 


PBBCENTAGB. 


7.  What  is  the  amount  of  $700  for  3  years  9  months  and 
24  days,  at  7  per  cent  compound  interest  ?    A  ns.  $906.55  + . 

A  more  expeditious  method  of  computing  compound,  in- 
terest than  the  preceding,  is  by  means  of  the  following 

Table, 

STwwing  the  amount  o/  $1,  <w  £1,  at  3,  4,  5,  6,  and  7  per  cent,  cott^ 

pound  interest,  for  any  number  of  years,  from  1  to  20. 


Yrs. 

3  per  cent 

4  per  cent. 

5  per  cent. 

6  per  cent. 

7  per  cent. 

1.. 

2.. 
3.. 
4.. 
5.. 

1.030,000 
1.060,900 
1.092,727 
1.125,509 
1.159,274 

1.040,000 
1.081,600 
1.124,864 
1.169,859 
1.216,653 

1.050,000 
1.102,500 
1.157,625 
1.215,506 

1.276,282 

1.060,000 
1.123,600 
1.191,016 
1.262,477 
1.338,226 

1.07,000 
1.14,49a 
1.22,504 
1.31,079 
1.40,255 

6.. 
7.. 
8.. 
9.. 
10.. 

1.194,052 
1.229,874 
1.266,770 
1.304,773 
1.343,916 

1.265,319 
1.315,932 
1.368,569 
1.423,312 
1.480,244 

1.340,096 
1.407,100 
1.477,455 
1.551,328 
1.628,895 

1.418,519 
1.503,630 
1.593,848 
1.689,479 
1.790,848 

1.50,073 
1.60,578 
1.71,818 
1.83,845 
1.96,715 

11.. 
12.. 
13.. 
14.. 
15.. 

1.384,234 
1.425,761 
1.468,534 
1.512,590 
1.557,967 

1.539,454 
1.601,032 
1.665,074 
1.731,676 
1.800,944 

1.710,339 
1.795,856 
1.885,649 
1.979,932 
2.078,928 

1.898,299 
2.012,196 
2.132,928 
2.260,904 
2.396,558 

2.10,485 
2.25,219 
2.40,984 
2.57,853 
2.75,903 

16.. 
17.. 

18.. 
19.. 
20.. 

1.604,706 
1.652,848 
1.702,433 
1.753,506 
1.806,111 

1.872,981 
1.947,900 
2.025,817 
2.106,849 
2.191,123 

2.182,875 
2.292,018 
2.406,619 
2.526,950 
2.653,298 

2.540,352 
2.692,773 
2.854,339 
3.025,600 
3.207,135 

2.95,216 
3.15,881 
3.37,293 
3.61,652 
3.86,968 

8.  What  is  the  amount  of  $800  for  6  years,  at  7  per  cent.  ? 

OPERATION. 

From  the  table    $1.50073        Amount  of  $1  for  the  time. 
800    Principal, 

11200.58400,  Ans. 

9.  What  is  the  compound  interest  of  $120  for  15  years,  at 
6  per  cent.  ?  A  ns.  $129.47  + . 

Of  what  use  is  the  table  in  computing  compound  interest  ? 


DISCOUNT.  259 

10.  What  is  the  amount  of  $.10  for  20  years,  at  7  per 
cent.  ?  Ans.  $.38696. 


DISGOUNT. 

325.  Discount  is  an  abatement  or  allowance  made  for 
the  payment  of  a  debt  before  it  is  due. 

326.  The  Present  Wortli  of  a  debt,  payable  at  a  future 
time  without  interest,  is  such  a  sum  as,  being  put  at  legal  in- 
terest, will  amount  to  the  given  debt  when  it  becomes  due. 

1.  A  owes  B  $321,  payable  in  1  year  ;  what  is  the  present 
worth  of  the  debt,  the  use  of  money  being  worth  7  per  cent.? 

OPERATION.  Analysis.  The 

Am't  of  $1     1.07  )  $321  (  $300,  Present  value,     amount  of  $1  for 

321  1  year  is  $1.07; 

tlierefore        tlie 

$321  Given  sum  or  debt.  present  wortli  of 

300  Present  wortli.  every    $1.07    of 

$21   Discount.  the  given  debt  is 

$1  ;  and  the  pres- 
ent worth  of  $321  will  be  as  many  dollars  as  $1.07  is  contained  times 
in  $331.    $831-r-1.07=$300,  Ans. 

Rule.  I.  Divide  the  given  sum  or  debt  ly  the  amount  of 
$1  for  the  given  rate  and  time,  and  the  quotient  will  he  the 
present  worth  of  the  delt. 

II.  SuUract  the  present  worth  from  the  given  sum  or  dedt, 
and  the  remainder  will  he  the  discount. 

The  terms  present  woi'th,  discount^  and  debt,  are  equivalent  to  2^nci2)al,  interest^ 
and  amount.  Hence,  when  the  time,  rate  per  cent,  and  amount  are  given,  the  prin- 
cipal may  he  found  by  (321) ;  and  the  interest  by  subtracting  the  principal  from 
the  amount. 

Examples  foe  Pkactice. 

2.  What  is  the  present  worth  of  $180,  payable  in  3  years 
4  months,  discounting  at  Q%  ?  Ans,  $150. 

Define  discount.    Present  worth.     Give  analysis.     Rule, 


260  PERCEKTAQE. 

3.  What  is  the  present  worth  of  a  note  for  11315.389,  due 
in  2  years  6  months,  at  7  per  cent.?  Ans.  $1119.48. 

4.  What  is  the  present  worth  of  a  note  for  $866,038,  due 
in  3  years  6  months  and  6  days,  when  money  is  worth  8  per 
cent.  ?    What  the  discount  ?        Ans.  1190.15 +  ,  discount. 

5.  What  is  the  present  worth  of  a  debt  for  $1005,  on 
which  $475  is  to  be  paid  in  10  months,  and  the  remainder 
in  1  year  3  months,  the  rate  of  interest 


being  e^^^J,   w^  , 

lout  interest,  find  tne  pres-     r 


When  payments  are  to  be  made  at  different  times  without 
ent  worth  of  each  payment  separately,  and  take  their  sum. 

6.  I  hold  a  note  against  C  for  $529,925,  due  Sept.  1, 
1859  ;  what  must  I  discount  for  the  payment  of  it  to-day, 
Feb.  7,  1859,  money  being  worth  6%?  Ans.  $17,425. 

7.  A  man  was  offered  $3675  in  cash  for  his  house,  or 
$4235  in  3  years,  without  interest ;  he  accepted  the  latter 
offer ;  how  much  did  he  lose,  money  being  worth  7  per 
cent.?  '  A?is.  $175. 

8.  A  man,  having  a  span  of  horses  for  sale,  offered  them 
for  $480  cash  in  hand,  or  a  note  of  $550  due  in  1  year  8 
months,  without  interest;  the  buyer  accepted  the  latter 
offer ;  did  the  seller  gain  or  lose  thereby,  and  how  much, 
interest  being  6%  ?  Ans.  Seller  gained  $20. 

9.  What  must  be  discounted  for  the  present  payment  of  a 
debt  of  $2637.72,  of  which  $517.50  is  to  be  paid  in  6  months, 
S 7 93. 75  in  10  months,  and  the  remainder  in  1  year  6  months, 
the  use  of  money  being  worth  7  per  cent.  ?   Ans.  $187.29  -f. 

10.  What  is  the  difference  between  the  interest  and  dis- 
count of  $130,  due  10  months  hence,  at  10;^  ?    A71S.  $.83 J. 

Promiscuous  Examples  m  Percentage. 
1.  A  merchant  bought  sugar  in  New  York  at  6 J-  cents 
per  pound  ;  the  wastage  by  transportation  and  retailing  was 
5  per  cent.,  and  the  interest  on  the  first  cost  to  the  time  of 
sale  was  2  per  cent. ;  how  much  must  he  ask  per  pound  to 
gain  25  per  cent.?  Ans.  8^-1-  cents. 


PROMISCUOUS    EXAMPLES.  261 

2.  A  person  purchased  2  lots  of  land  for  $200  eacli,  and 
sold  one  at  40  per  cent,  more  than  cost,  and  the  other  at  20 
per  cent,  less ;  what  was  his  gain  ?  Ans.  $40. 

3.  Sold  goods  to  the  amount  of  $425,  on  6  months'  credit, 
which  was  125  more  than  the  goods  cost ;  what  was  the  true 
profit,  money  being  worth  6%  ?  A71S.  112. 62  + . 

4.  Bought  cotton  cloth  at  13  cents  a  yard,  on  8  months' 
credit,  and  sold  it  the  same  day  at  12  cents  casJi ;  how  much 
did  I  gain  or  lose  per  cent. ,  money  being  worth  6  per  cent.? 

5.  A  farmer  sold  a  pair  of  horses  for  $150  each  ;  on  one 
he  gained  25  per  cent.,  on  the  other  he  lost  25  per  cent.;  did 
he  gain  or  lose  on  both,  and  how  much  ?    Ans.  Lost  $20. 

6.  A  man  inyested  f  of  all  he  was  worth  in  the  coal  trade, 
and  at  the  end  of  2  years  8  months  sold  out  his  entire  inter- 
est for  $3100,  which  was  a  yearly  gain  of  9  per  cent,  on  the 
money  invested ;  how  much  was  he  worth  when  he  com- 
menced trade  ?  Ans.  $3750. 

7.  In  how  many  years  will  a  man,  paying  interest  at  7 
per  cent,  on  a  debt  for  land,  pay  the  face  of  the  debt  in 
interest  ?  Ans.  14f  years. 

8.  Two  persons  engaged  in  trade  ;  A  furnished  |-  of  the 
capital,  and  B  | ;  and  at  the  end  of  3  years  4  months  they 
found  they  had  made  a  clear  profit  of  $5000,  which  was  12| 
per  cent,  per  annum  on  the  money  invested  ;  how  much  cap- 
ital did  each  furnish?  Ans.  A,  $7500;  B,  $4500. 

^9.  Bought  $500  worth  of  dry  goods,  and  $800  worth  of 
groceries  ;  on  the  dry  goods  I  lost  20  per  cent.,  but  on  the 
groceries  I  gained  15  per  cent.;  did  I  gain  or  lose  on  the 
whole  investment,  and  how  much  ?         Ans.  Gained  $20. 

10.  What  amount  of  accounts  must  an  attorney  collect, 
in  order  to  pay  over  $1100,  and  retain  8-^  per  cent,  for  col- 
lecting? Ans.U200 

11.  A  merchant  sold  goods  to  the  amount  of  $667,  to  be 
paid  in  8  months ;  the  same  goods  cost  him  $600  one  year 
previous  to  the  sale  of  them  ;  money  being  worth  6  per  cent., 
what  was  his  true  gain?  Ans.  $5,346  +  . 


262  PERCENTAGE. 

12.  A  nurseiyman  sold  trees  at  $18  per  hundred,  and 
cleared  J  of  his  receipts  ;  what  per  cent,  profit  did  he  make  ? 

Ans.  50^. 

13.  If  f  of  an  article  be  sold  for  what  -f  of  it  cost,  what  is 
the  gain  per  cent.  ?  Ans.  40|;^. 

14.  A  lumber  merchant  sells  a  lot  of  lumber,  which  he  has 
had  on  hand  6  months,  on  10  months'  credit,  at  an  advance  of 
30  per  cent,  on  the  first  cost ;  if  he  is  paying  5  per  cent,  inter- 
est on  capital,  what  are  his  profits  per  cent.?  Ans.  21-J^. 

15.  A  person,  owning  -f  of  a  piece  of  property,  sold  20  per 
cent,  of  his  share  ;  what  part  did  he  then  own?     Ans.  J. 

16.  A  speculator,  having  money  in  the  bank,  drew  60  per 
cent,  of  it,  and  expended  30  per  cent,  of  50  per  cent,  of  this 
for  728  bushels  of  wheat,  at  |1.12|-  per  bushel ;  how  much 
was  left  in  the  bank  ?  Ans.  ^SQAO. 

17.  I  wish  to  line  the  carpet  of  a  room,  that  is  6  yards 
long  and  5  yards  wide,  with  duck  f  yard  wide ;  how  many 
yards  of  lining  must  I  purchase,  if  it  will  shrink  4  per  cent, 
in  length,  and  5  per  cent,  in  width  ?  Ans.  43f  f . 

18.  A's  money  is  28  per  cent,  more  than  B's ;  how  many 
per  cent,  is  B's  less  than  A's  ?  A7ts.  21^^. 

19.  A  capitalist  invested  f  of  his  money  in  railroad  stock, 
which  depreciated  5  per  cent,  in  value  ;  the  remaining  f  he 
invested  in  bank  stock,  which,  at  the  end  of  1  year,  had 
gained  $1200,  which  was  12  per  cent,  of  the  investment ; 
what  was  the  whole  amount  of  his  capital,  and  what  was  his 
entire  loss  or  gain  ?       Ans.  $25000,  capital ;  $450,  gain. 

20.  O's  money  is  to  D's  as  2  to  3  ;  if  I  of  C's  money  be 
put  at  interest  for  3  years  9  months,  at  10  per  cent.,  it  will 
amount  to  $1933.25  ;  how  much  money  has  each  ? 

BANKING. 
327.  A  Bank  is  a  corporation  chartered  by  law  for  the 

purpose  of  receiving  and  loaning  money,  and  furnisliiug  a 
paper  circulation. 

What  is  a  bank  ? 


BAlfKING.  263 

328.  A  Promissory  Note  is  a  -WTitteii  or  printed  en- 
gagement to  pay  a  certain  sum,  either  on  demand  or  at  a 
specified  time . 

329.  Bank  Notes,  or  Bank  Bills,  are  the  notes  made 
and  issued  by  banks  to  circulate  as  money.  They  are  pay- 
able in  specie  at  the  banks. 

330.  The  Face  of  a  note  is  the  sum  made  payable  by 
the  note. 

331.  Days  of  Grace  are  the  three  days  usually  allowed 
by  law  for  the  payment  of  a  note  after  the  expiration  of  the 
time  specified  in  the  note. 

332.  The  Maturity  of  a  note  is  the  expiration  of  the 
days  of  grace  ;  a  note  is  due  at  maturity. 

333.  Notes  may  contain  a  promise  of  interest,  which  will 
be  reckoned  from  the  date  of  the  note,  unless  some  other 
time  be  specified. 

The  transaction  of  borrowing  money  at  banks  is  conducted 
in  accordance  with  the  following  custom  :  the  borrower  pre- 
sents a  note,  either  made  or  indorsed  by  himself,  payable  at 
a  specified  time,  and  receives  for  it  a  sum  equal  to  the  face, 
less  the  interest  for  the  time  the  note  has  to  run.  The 
amount  thus  withheld  by  the  bank  is  in  consideration  of 
advancing  money  on  the  note  prior  to  its  maturity. 

334.  Bank  Discount  is  an  allowance  made  to  a  bank 
for  the  payment  of  a  note  before  it  becomes  due. 

335*  The  Proceeds  of  a  note  is  the  sum  received  for  it 
when  discounted,  and  is  equal  to  the  face  of  the  note  less 
the  discount. 

Case  L 
336.  Given  the  face  of  a  note  to  find  the  proceeds. 
The  law  of  custom  at  banks  makes  the  discount  of  a  note 

Define  a  promissory  note.  Bank  notes.  The  face  of  a  note.  Days 
of  grace.  The  maturity  of  a  note.  Explain  the  process  of  discount- 
ing a  note  at  a  bank.  Define  bank  discount.  The  proceeds  of  a  note. 
W)jatisCaseI? 


264  PEECEKTAGE. 

equal  to  the  simple  interest  at  the  legal  rate  for  the  time 
specified  in  the  note. 

Rule.  I.  Compute  the  interest  on  the  face  of  the  note  for 
three  days  more  than  the  specified  time  ;  the  result  will  be  the 
discount. 

11.  Subtract  the  discount  from  the  face  ofthenote^  and  the 
remainder  will  le  the  proceeds. 

Examples  for  Practice. 

1.  What  is  the  discount,  and  what  the  proceeds,  of  a  note 
for  $450,  at  60  days,  discounted  at  a  bank  at  Q%  ? 

Ans.  Discount,  $4,725  ;  proceeds,  $445,275. 

2.  What  are  the  proceeds  of  a  note  for  $368,  at  90  days, 
discounted  at  the  Bank  of  New  York?   Ans.  $361,345  +  . 

3.  What  shall  I  receive  on  my  note  for  $475.50,  at  60 
days,  if  discounted  at  the  Crescent  City  Bank,  New  Or- 
leans? ^W5.  $471.33  +  . 

4.  What  are  the  proceeds  of  a  note  for  $10000,  at  90 
days,  discounted  at  the  Philadelphia  Bank  ?  Ans.  $9845. 

6.  Paid,  in  cash,  $240  for  a  lot  of  merchandise.  Sold  it 
the  same  day,  receiving  a  note  for  $250  at  60  days,  which  I 
got  discounted  at  the  Hartford  Bank.  'What  did  I  make 
by  this  speculation  ?  Ans.  $7.37|-. 

6.  A  note  for  $360.76,  drawn  at  90  days,  is  discounted  at 
the  Vermont  Bank.     Find  the  proceeds.    ^W5.  $355,168  +  . 

7.  Wishing  to  borrow  $530  of  a  western  bank  which  is 
discounting  paper  at  8  per  cent.,  I  give  my  note  for  $536.75, 
payable  in  60  days.  How  much  do  I  need  to  make  up  the 
required  amount  ?  Ans.  $.7645. 

1.  To  indicate  the  maturity  of  a  note  or  draft,  a  vertical  line  ( I )  is  need,  with  the 
day  at  which  the  note  is  nominally  due  on  the  left,  and  the  date  of  maturity  on  the 
right ;  thus,  Jan.  '  1 1  o- 

2.  When  a  note  is  on  inter tst^  payable  at  a  ftiture  specified  time,  the  amount  is  the 
foce  of  the  note,  or  the  sum  made  payable,  and  must  be  made  the  basis  of  discount. 

Give  rule. 


B  A  IT  K I  ]sr  G .  265 

Find  the  maturity,  term  of  discount,  and  proceeds  of  the 
following  notes : 

^500.  BosTOK,  Jan.  4,  1859. 

8.  Three  months  after  date,  I  promise  to  pay  to  the  order 
of  John  Brown  &  Co.  five  hundred  dollars,  at  the  Suffolk 
Bank,  value  received.  James  Barker. 

Discounted  March  2.  f  Due,  April*],. 

Ans,   -l  Term  of  discount,  36  da. 

I  Proceeds,  $497. 

^^50.  St.  Louis,  June  12, 1859. 

9.  Six  months  after  date,  I  promise  to  pay  Thomas  Lee, 
or  order,  seven  hundred  fifty  dollars,  with  interest,  value 
received.  Bykok  QuiiTBY. 

Discounted  at  a  broker's,  Nov.  15,  at  10^. 

rDue,  Dec.  i2|i5. 

Ans,   -l  Term  of  discount,  30  da. 

L  Proceeds,       $766,434+. 

Case  IL 

337.  G-iven  the  proceeds  of  a  note,  to  find  the 
face. 

1.  I  wish  to  borrow  $400  at  a  bank.  For  what  sum  must 
I  draw  my  note,  payable  in  60  days,  so  that  when  discounted 
at  6  per  cent.  I  shall  receive  the  desired  amount  ? 

OPERATION.  Analysis.     $400  is  the 

$1.0000  proceeds  of  a  certain  note, 

.0105  =  disc,  on  $1  for  63  da.  *^e  ^8-ce  of  which  we  are 

required  to  find.    We  first 

$  .9895  =  proceeds  of  |1.  ^^t^in  the  proceeds  of  $1 

$400  -T-  .9895  =  $404,244  =  by  the  last  case,  and  then 

face  of  the  required  note.  divide  the  given  proceeds, 

$400,  by  this  sum;  for,  as  many  times  as  the  proceeds  of  $1  is  con- 
tained in  the  given  proceeds,  so  many  dollars  must  be  the  face  of  the 
required  note. 

Give  Case  II,    Analysis. 
E.P.  13 


HGG  PEBCENTAOE. 

Rule.  Divide  the  proceeds  ly  tlie  proceeds  of  $1  for  the 
time  and  rate  mentioned,  and  the  quotient  will  be  the  face  of 
the  note. 

Examples  fob  Pbactice. 

2.  What  is  the  face  of  a  note  at  60  days,  which  yields 
$680  when  discounted  at  a  New  Hayen  bank  ? 

Ans.  $687,215. 

3.  What  is  the  face  of  a  note  at  90  days,  of  which  the 
proceeds  are  $1000  when  discounted  at  a  Louisiana  bank  ? 

Ans,  $1013.085. 

4.  Wishing  to  borrow  $500  at  a  bank,  for  what  sum  must 
my  note  be  drawn,  at  30  days,  to  obtain  the  required  amount, 
discount  being  at  1%  ?  Ans,  $503.22. 

5.  James  Hopkins  buys  merchandise  of  me  in  New  York, 
at  cash  price,  to  the  amount  of  $1256.  Not  having  money, 
he  gives  his  note  in  payment,  drawn  at  6  months.  What 
must  be  the  face  of  the  note  ?  Ans.  $1302.341. 

EXCHANGE. 

338.  Exchange  is  a  method  of  remitting  money  from 
one  place  to  another,  or  of  making  payments  by  written 
orders. 

339.  A  Bill  of  Exchange  is  a  written  request  or  order 
upon  one  person  to  pay  a  certain  sum  to  another  person,  or 
to  his  order,  at  a  specified  time. 

340.  A  Sight  Draft  or  Bill  is  one  requiring  payment  to 
be  made  "  at  sight,"  which  means,  at  the  time  of  its  presenta- 
tion to  the  person  ordered  to  pay.  In  other  bills,  the  time 
specified  is  usually  a  certain  number  of  days  *' after  sight." 

There  are  always  three  parties,  and  usually  four,  to  .^ 
transaction  in  exchange. 

341.  The  Drawer  or  Maker  is  the  person  who  signs 
the  order  or  bill. 

Give  the  rule.  Define  exchange.  A  hill  of  exchange.  A  sight 
draft.    The  drawer. 


EXCHANGE.  267 

342.  The  Drawee  is  the  person  to  whom  the  order  is 
addressed. 

343.  The  Payee  is  the  person  to  whom  the  money  is 
ordered  to  be  paid. 

344.  The  Buyer  or  Remitter  is  the  person  who  pur- 
chases the  bill.  He  may  be  himself  the  payee,  or  the  bill 
may  be  drawn  in  favor  of  any  other  person. 

345.  The  Indorsement  of  a  bill  is  the  writing  upon  its 
back,  by  which  i\iQ  payee  relinquishes  his  title,  and  transfers 
the  payment  to  another.  The  payee  may  indorse  in  Uanh 
by  writing  his  name  only,  which  makes  the  bill  payable  to 
the  hearer,  and  consequently  transferable  like  a  bank  note ; 
or  he  may  accompany  his  signature  by  a  special  order  to  pay 
to  another  person,  who  in  his  turn  may  transfer  the  title  in 
like  manner,  Indorsers  become  separately  responsible  for 
the  amount  of  the  bill,  in  case  the  drawee  fails  to  make  pay- 
ment. A  bill  made  payable  to  the  dearer  is  transferable 
without  indorsement. 

346.  The  Acceptance  of  a  bill  is  the  promise  which  the 
draivee  makes  when  the  bill  is  presented  to  him  to  pay  it  at 
maturity ;  this  obligation  is  usually  acknowledged  by  writ- 
ing the  word  "Accepted,"  with  his  signature,  across  the 
face  of  the  bill. 

Three  days  of  grace  are  usually  allowed  for  the  payment  of  a  hill  of  exchange 
after  the  time  specified  has  expired.  But  in  New  York  State  no  grace  is  allowed  on 
sight  drafts. 

From  these  definitions,  the  use  of  a  bill  of  exchange  in 
monetary  transactions  is  readily  perceived.  If  a  man  wishes 
to  make  a  remittance  to  a  creditor,  agent,  or  any  other  per- 
son residing  at  a  distance,  instead  of  transporting  specie, 
which  is  attended  with  expense  and  risk,  or  sending  bank 
notes,  which  are  liable  to  be  uncurrent  at  a  distance  from 
the  banks  that  issue  them,  he  remits  a  bill  of  exchange, 
purchased  at  a  bank  or  elsewhere,  and  made  payable  to  the 

The  drawee.  The  payee.  The  buyer.  An  indorsement.  An 
acceptance.    What  of  grace  on  bills  of  exchange  ? 


268  PEECENTAGE. 

proper  person  in  or  near  the  place  where  he  resides.  Thus 
a  man  by  paying  Boston  funds  in  Boston,  may  put  New- 
York  funds  into  the  hands  of  his  New  York  agent. 

347.  The  Course  of  Exchange  is  the  variation  of  the 
cost  of  sight  hills  from  their  par  value,  as  affected  by  the  rela- 
tive conditions  of  trade  and  commercial  credit  at  the  two 
places,  between  which  exchange  is  made.  It  may  be  either  at 
a  premium  or  discount,  and  is  rated  at  a  certain  per  cent,  on 
the  face  of  the  bill.  Bills  payable  a  specified  time  after  sight 
are  subject  to  discount,  like  notes  of  hand,  for  the  term  of 
credit  given.  Hence  their  value  in  the  money  market  is 
affected  by  both  the  course  of  exchange  and  the  discount 
for  time. 

348.  Foreign  Exchange  relates  to  remittances  made 
between  different  countries. 

349.  Domestic  or  Inland  Exchange  relates  to  remit- 
tances made  between  different  places  in  the  same  country. 

An  inland  bill  of  exchange  is  commonly  called  a  Draft. 
In  this  work  we  shall  treat  only  of  Inland  Exchange, 

Case  I. 

350.  To  find  the  cost  of  a  draft. 


^500.  Syracuse,  May  7,  1859. 

1.  At  sight,  pay  to  James  Clark,  or  order,  five  hundred 
dollars,  value  received,  and  charge  the  same  to  our  account. 

M.  Smith  &  Co. 
To  Messrs.  Brown  &  Foster,  ) 
Baltimore.         ) 
What  is  the  cost  of  the  above  draft,  the  rate  of  exchange 
being  1 J  per  cent,  premium  ? 

OPERATION.  Analysis.    Since  ex- 

$500  X  1.015  =  $507.50,  Ans.  change  is  at  1^  per  cent. 

premium,  each  dollar  of 
the  draft  will  cost  $1,015;  and  to  find  the  whole  cost  of  the  draft. 

How  is  exchange  conducted  ?  Explain  course  of  exchange.  For- 
eign exchange.  Inland  exchange.  Define  a  draft.  What  is  Case  I  ? 
Give  analysis. 


EXCHAKGE.  269 

we  multiply  its  face,  $500,  by  1.015,  and  obtain  $507.50,  tbe  required 
result. 


BoSTOK,  June  12,  1859. 


2.  Thirty  days  after  sight,  pay  to  John  Otis,  or  bearer, 
four  hundred  eighty  dollars,  value  received,  and  charge  the 
same  to  account  of  Amos  Teenchakd. 

To  John  Stiles  &  Co., 
New  York. 


What  is  the  cost  of  the  above  draft,  exchange  being  at  a 
premium  of  S%  ? 

OPEEiATiON.  Analysis.    Since  time 

$1.0000  is  allowed,  the  draft  must 

.0055  =  discount  for  33  days.  Suffer  discount  in  the  sale. 

The  discount  of  $1,  at  the 

$  .9945  =  proceeds  of  $1.  je^al  rate  in  Boston,  for 

.03       rz:  rate  of  exchange.  the    specified    time,    al- 

$1.0245  =  cost  of  $1  of  the  draft  ^^^^^  g^^^e,  is   $.0055, 

$480  X  1.0245  =  $491.76,  Ans.        j;^^"^^'  ^TZT^  ^'°°' 

$1,  gives  $.9945,  the  cost 

of  $1  of  the  draft,  provided  sight  exchange  were  at  par  ;  but  sight  ex- 
change being  at  premium,  we  add  the  rate,  .03,  to  .9945,  and  obtain 
$1.0245,  the  actual  cost  of  $1.  Then,  multiplying  $480  by  1.0245,  we 
obtain  $491.76. 

EuLE.  I.  For  sight  drafts. — Multiply  the  face  of  the  draft 
dy  1  plus  the  rate  when  exchange  is  at  a  premium,  and  by 
1  minus  the  rate  when  exchange  is  at  a  discount, 

II.  For  drafts  payable  after  sight. — Find  the  proceeds  of  $1 
at  lank  discount  for  tlie  specified  time,  at  the  legal  rate  where 
the  draft  is  purchased;  then  add  the  rate  of  exchange  when 
at  a  premium,  or  subtract  it  when  at  a  discount,  and  multiply 
the  face  of  the  draft  by  this  result. 

Examples  fou  Practice. 
3.  A  merchant  in  Cincinnati  wishes  to  remit  $1000  by 
Give  analysis.    Rrle  I ;  II. 


270  PERCENTAGE. 

draft  to  his  agent  in  New  York ;  what  will  the  bill  cost,  ex- 
change being  at  3  per  cent,  premium  ?  Ans.  $1030. 

4.  What  will  be  the  cost  in  Rochester  of  a  draft  on  Albany 
for  $400,  payable  at  sight,  exchange  being  at  |  per  cent, 
premium?  A^is.  $403. 

5.  A  merchant  m  St.  Louis  orders  goods  from  New  York, 
to  the  amount  of  $530,  which  amount  he  remits  by  draft, 
exchange  being  at  2}  per  cent,  premium.  If  he  pays  $20  for 
transportation,  what  will  the  goods  cost  him  in  St.  Louis  ? 

Ans.  $564,575. 

6.  What  will  be  the  cost,  in  Detroit,  of  a  draft  on  Boston 
for  $800,  payable  60  days  after  sight,  exchange  being  at  a 
premium  of  2%  ?  Ans.  $806.20. 

7.  A  man  in  Philadelphia  purchased  a  draft  on  Chicago  for 
$420,  payable  30  days  after  sight ;  what  did  it  cost  him,  the 
rate  of  exchange  being  1 J  per  cent,  discount  ?    Ans.  $411.39. 

8.  A  merchant  in  Portland  receives  from  his  agent  320 
barrels  of  flour,  purchased  in  Chicago  at  $10  per  barrel ;  in 
payment  for  which  he  remits  a  draft  on  Chicago,  at  2^  per 
cent,  discount.  The  transportation  of  his  flour  cost  $312. 
What  must  he  sell  it  for  per  barrel  to  gain  $400  ? 

Ans.  $12. 

Case  IL 

351.  To  find  the  face  of  a  draft  which  a  given 
sum  will  purchase. 

1.  A  man  in  Indiana  paid  $369.72  for  a  draft  on  Boston, 
drawn  at  30  days  ;  what  was  the  face  of  the  draft,  exchange 
being  at  3^  per  cent,  premium  ? 

OPERATION.  Analysis.    We  find, 

$369.72  -T-  1.027  =  $360,  Ans.  hy  Case  I,  that  a  draft 

for  $1  will  cost  $1,027; 
hence  the  draft  that  will  cost  $369.72  must  be  for  as  many  dollars  as 
1.027  is  contained  times  in  $369.72;  dividing,  we  obtain  $360,  the 
required  result. 

What  is  Case  II  ?    Give  analysis. 


EQUATION    OF    PAYMENTS.  271 

Rule.  Divide  the  given  cost  hy  the  cost  of  a  draft  for  II, 
at  the  given  rate  of  exchange ;  the  quotient  will  be  the  face 
of  the  required  draft. 

Examples  for  Practice. 

2.  What  draft  may  be  purchased  for  $243.60,  exchange 
being  at  H  per  cent,  premium?  Ans.  1240. 

3.  What  draft  may  be  purchased  for  $79.20,  exchange  be- 
ing at  1  per  cent,  discount  ?  Ans,  $80. 

4.  An  agent  in  Pittsburg  holding  $282.66,  due  his  em- 
ployer in  New  Haven,  is  directed  to  make  the  remittance 
by  draft,  drawn  at  60  days.  What  will  be  the  face  of  the 
draft,  exchange  being  at  2  per  cent,  premium  ?    Ans.  $280. 

6.  An  emigrant  from  Bangor  takes  $240  in  bank  bills  to 
St.  Paul,  Minn.,  and  there  pays  ^  per  cent,  brokerage  in  ex- 
change for  current  money.  What  would  he  have  saved  by 
purchasing  in  Bangor  a  draft  on  St.  Paul,  drawn  at  30  days, 
exchange  being  at  IJ  per  cent,  discount?       Ans.  $5,599. 

6.  A  Philadelphia  manufacturer  is  informed  by  his  agent 
in  Buffalo  that  $3600  is  due  him  on  the  sale  of  some  prop- 
erty. He  instructs  the  agent  to  remit  by  a  draft  payable  in 
60  days  after  sight,  exchange  being  at  |  per  cent,  premium. 
The  agent,  by  mistake,  remits  a  sight  draft,  which,  when 
received  in  Philadelphia,  is  accepted,  and  paid  after  the  ex- 
piration of  the  three  days  of  grace.  If  the  manufacturer 
immediately  puts  this  money  at  interest  at  the  legal  rate, 
will  he  gain  or  lose  by  the  blunder  of  his  agent  ? 

Ans.  He  will  lose  $8.24 -f-. 

EQUATION    OF    PAYMENTS. 

352.  Equation  of  Payments  is  the  process  of  finding 
the  mean  or  equitable  time  of  payment  of  several  sums,  due 
at  different  times  without  interest. 

353.  The  Term  of  Credit  is  the  time  to  elapse  before  a 
debt  becomes  due. 


Rule.     Define  equation  of  payments.     Term  of  credit. 


272  EQUATION    OF    PAYMENTS. 

354.  The  Average  Term  of  Credit  is  the  time  to  elapse 
before  several  debts,  due  at  different  times,  may  all  be  paid 
at  once,  without  loss  to  debtor  or  creditor. 

355.  The  Equated  Time  is  the  date  at  which  the  sev- 
eral debts  may  be  canceled  by  one  payment. 

Case  I. 

356.  When  all  the  terms  of  credit  begin  at  the 
same  date. 

1.  On  the  first  day  of  January  I  find  that  I  owe  Mr. 
Smith  8  dollars,  to  be  paid  in  5  months,  10  dollars  to  be 
paid  in  2  months,  and  12  dollars  to  be  paid  in  10  months; 
at  what  time  may  I  pay  the  whole  amount  ? 

OPEKATION. 

$  8  X  5  =  40 
10  X  2  =  20 
12  X  10  =  120 

30  180  -^  30  r=  6  mo.,  average  time  of  credit. 

Jan.  1+6  mo.  =  July  1,  equated  time  of  payment 

Analysis.  The  whole  amount  to  be  paid,  as  seen  above,  is  $30 ; 
and  we  are  to  find  how  long  it  shall  be  withheld,  or  what  term  of 
credit  it  shall  have,  as  an  equivalent  for  the  various  terms  of  credit 
on  the  different  items.  Now,  the  value  of  credit  on  any  sum  is  meas- 
ured by  the  product  of  the  money  and  time.  And  we  say,  the  credit 
on  $8  for  5  mo.  =  the  credit  on  $40  for  1  mo.,  because  8  x  5  =  40  x  1. 
In  the  same  manner,  we  have,  the  credit  on  $10  for  2  mo.  =:  the  credit 
on  $20  for  1  mo. ;  and  the  credit  on  $12  for  10  mo.  =  the  credit  on  $120 
for  1  mo.  Hence,  by  addition,  the  value  of  the  several  terms  of  credit 
on  their  respective  sums  equals  a  credit  of  1  month  on  $180 ;  and  this 
equals  a  credit  of  6  months  on  $30,  because 
30  X  6  =  180  X  1. 

Rule.  L  Multiply  each  payment  iy  its  term  of  credit, 
and  divide  the  sum  of  the  products  by  the  sum  of  the  pay- 
ments ;  the  quotient  will  be  the  average  term  of  credit. 

Average  term  of  credit.  Equated  time.  Give  Case  I.  Analysis. 
Rule. 


AYERAGIKG    CREDITS.  273 

II.  Add  tTie  average  term  of  credit  to  the  date  at  which  all 
the  credits  legin,  and  the  result  will  be  the  equated  time  of 
payment, 

1.  The  periods  of  time  used  as  multipliers  must  all  be  of  the  same  denomination, 
and  the  quotient  will  be  of  the  same  denomination  as  the  terms  of  credit ;  if  these 
be  months,  and  there  be  a  remainder  after  the  division,  continue  the  division  to 
days  by  reduction,  always  taking  the  nearest  unit  in  the  last  result. 

2.  The  several  rules  in  equation  of  payments  are  based  upon  the  principle  of  bank 
discount :  for  they  imply  that  the  discount  of  a  sum  paid  before  it  is  due  equals  the 
Interest  of  the  same  amount  paid  after  it  is  due. 

Examples  foe  Peactice. 

2.  On  the  25th  of  September  a  trader  bought  merchandise, 
as  follows  :  $700  on  20  days'  credit ;  1400  on  30  days'  credit; 
$700  on  40  days'  credit :  what  was  the  average  term  of  credit, 
9,nd  what  the  equated  time  of  payment  ? 

,       ( Average  credit,  30  days. 

*  1  Equated  time  of  payment,  Oct.  25. 

3.  On  July  1  a  merchant  gave  notes,  as  follows  :  the  first 
for  $250,  due  in  4  months ;  the  second  for  $750,  due  in  2 
months  ;  the  third  for  $500,  due  in  7  months  :  at  what  time 
may  they  all  be  paid  in  one  sum  ?  Ans,  Nov.  1. 

4.  A  farmer  bought  a  cow,  and  agreed  to  pay  $1  on  Mon- 
day, $2  on  Tuesday,  $3  on  Wednesday,  and  so  on  for  a  week ; 
desirous  afterward  to  avoid  the  Sunday  payment,  he  offered 
to  pay  the  whole  at  one  time  :  on  what  day  of  the  week 
would  this  payment  come  ?  Ans.  Friday. 

5.  Jan.  1, 1  find  myself  indebted  to  John  Kennedy  in  sums 
as  follows  :  $650  due  in  4  months  ;  $725  due  in  8  months ;  and 
$500  due  in  12  months  :  at  what  date  may  I  settle  by  giving 
my  note  on  interest  for  the  whole  amount  ?    Ans.  Aug.  21. 

Case  II. 

357.  "When  the  terms  of  credit  begin  at  different 
dates,  and  the  acconnt  has  only  one  side. 

358.  An  Account  is  the  statement  or  record  of  mercan- 
tile transactions  in  business  fonn. 

Give  Case  II.    Define  an  account. 
12* 


274 


EQUATIOl^    OF    PAYMENTS. 


359.  The  Items  of  an  account  may  be  sums  due  at  the 
date  of  the  transaction,  or  on  credit  for  a  specified  time. 

An  account  may  have  both  a  debit  and  a  credit  side,  the 
former  marked  Dr.,  the  latter  Or.  Suppose  A  and  B  have 
dealings  in  which  there  is  an  interchange  of  money  or  prop- 
erty ;  A  keeps  the  account,  heading  it  with  B's  name  ;  the 
Dr.  side  of  the  account  shows  what  B  has  received  from  A  ; 
the  Cr.  side  shows  what  he  has  parted  with  to  A. 

360.  The  Balance  of  account  is  the  difference  of  the 
two  sides,  and  may  be  in  favor  of  either  party. 

If,  in  the  transactions,  one  party  has  received  nothing  from 
the  other,  the  balance  is  simply  the  whole  amount,  and  the 
account  has  but  one  side.    Bills  of  purchase  are  of  this  class. 

Book  accounts  bear  interest  after  the  expiration  of  the  term  of  credit,  and  notes 
after  they  become  due. 

361.  To  Averagre  an  Account  is  to  find  the  mean  or 
equitable  time  of  payment  of  the  balance, 

362.  A  Focal  Date  is  a  date  to  which  all  the  others  are 
compared  in  averaging  an  account. 

1.  When  does  the  amount  of  the  following  bill  become 
due,  by  averaging  ? 
J.  0.  Smith, 

1859.  To  C.  E.  BoRDEif,        Dr. 

June  1.    To  Cash $450 

''  12.     ''  Mdse.  on4mos 500 

Aug.  16.     "  Mdse 250 


FIKST  OPERATION. 

SECOND  OPERATION. 

Due. 

da. 

Items. 

Prod. 

Due. 

da. 

Items. 

Prod. 

June    1 
Oct.    12 
Aug.  16 

0 

133 

76 

450 
500 
250 

66500 
19000 

June    1 
Oct.   12 

Aug.  16 

133 

0 

57 

450 
500 
250 

59850 
14250 

1200 

85500 

1200 

74100 

85500 -J- 1200  =  71  da. 
.        ( 71  da.  after  June  1, 
^»'- lor  Aug.  11. 


Atis. 


74100  -4- 1200  =  62  da. 
62  da.  before  Oct.  12, 
or  Aug.  11. 


Define  items.    Balance.    To  average  an  account.    A  focal  date. 


AVERAGING     ACCOUNTS.  275 

Analysis.  By  reference  to  the  example,  it  will  be  seen  that  the 
items  are  due  June  1,  Oct.  13,  and  Aug.  16,  as  shown  in  the  two  opera- 
tions. In  the  first  operation  we  use  the  earliest  date,  June  1,  as  a  focal 
date,  and  find  the  difference  in  days  between  this  date  and  each  of  the 
others,  regard  being  had  to  the  number  of  days  in  calendar  months. 
From  June  1  to  Oct.  12  is  133  da. ;  from  June  1  to  Aug.  16  is  76  da. 
Hence  the  firat  item  has  no  credit  from  June  1,  the  second  item  has 
133  days'  credit  from  June  1,  and  the  third  item  has  76  days'  credit 
from  June  1,  as  appears  in  the  column  marked  da.  After  this  we  pro- 
ceed precisely  as  in  Case  I,  and  find  the  average  credit,  71  da.,  and 
the  equated  time,  Aug.  11. 

In  the  second  operation,  the  latest  date,  Oct.  12,  is  taken  for  a  focal 
date ;  the  work  is  explained  thus :  Suppose  the  account  to  be  settled 
Oct.  12.  At  that  time  the  first  item  has  been  due  133  days,  and  must 
therefore  draw  interest  for  this  time.  But  interest  on  $450  for  133 
days  =  the  interest  on  $59850  for  1  da.  The  second  item  draws  no 
interest,  because  it  falls  due  Oct.  12.  The  third  item  must  draw  inter- 
est 57  days.  But  interest  on  $250  for  57  days  =  the  interest  on  $14250 
for  1  day.  Taking  the  sum  of  the  products,  we  find  the  whole  amount 
of  interest  due  on  the  account,  at  Oct.  12,  equals  the  interest  on  $74100 
for  1  day ;  and  this,  by  division,  is  found  to  be  equal  to  the  interest  on 
$1200  for  62  days,  which  time  is  the  average  term  of  interest.  Hence 
the  account  would  be  settled  Oct.  12,  by  paying  $1200  with  interest  on 
the  same  for  62  days.  This  shows  that  $1200  has  been  due  62  days  ; 
that  is,  it  falls  due  Aug.  11,  without  interest. 

EuLE.  I.  Find  the  time  at  tvhich  each  item  becomes  due, 
ly  adding  to  the  date  of  each  transaction  the  term  of  credit, 
if  any  he  s^iecified,  and  write  these  dates  in  a  column. 

II.  Assume  either  the  earliest  or  the  latest  date  for  a  focal 
date,  and  find  the  difference  in  days  between  the  focal  date 
and  each  of  the  other  dates,  and  write  the  results  in  a  second 
column. 

III.  Write  the  items  of  the  account  in  a  third  column,  and 
multiply  each  sum  by  the  corresponding  member  of  days  in  the 
preceding  column,  ivriting  the  products  in  a  final  column. 

IV.  Divide  the  sum  of  the  products  by  the  sum  of  the  items. 
The  quotient  will  be  the  average  term  of  credit  when   the 

Give  analysis.     Rule. 


276  EQUATION    OF    PAi-ME]S"TS. 

earliest  date  is  the  focal  date,  or  the  average  term  of  interest 
when  the  latest  date  is  the  focal  date  j  in  either  case  ahvays 
reckon,  from  the  focal  date  toward  the  other  dates,  to  find  the 
equated  time  of  payment 

Examples  for  Pkactice. 

2.  John  Brown, 

1859.  To  James  Greigg,     Dr. 

Jan.      1.    To    50  yds.  Broadcloth,  @  $3.00 ^150 

''      16.      '^2000    *^   Calico,  ''      .10 200 

Feb.     4.     ''      75    ''   Carpeting,     ''    1.33^ 100 

March  3.     ''   400    ''   Oil  Cloth,      ''      .40 160 

If  James  Greigg  wishes  to  settle  the  above  bill  by  giving 
his  note,  from  what  date  shall  the  note  draw  interest  ? 

Ans,  Jan.  27. 

3.  Abram  Russel, 

1859.  To  Wynkoop  &  Bro.,     Dr. 

March  1.     To  Cash $300 

April    4.       ''  Mdse 240 

June  18.       ''      ''      on  2  mo 100 

Aug.     8.      "  Cash 400 

What  is  the  equated  time  of  payment  of  the  above  account  ? 

Ans.  May  26. 

4.  John  Otis, 

1858.  To  James  Ladd,  Dr. 

June   1.  To  500  bu.  Wheat,  @  $1.20 $600 

''     12.  ''200''        "       "     1.50 300 

"     15.  "640"        "       "     1.30 832 

"     25.  "  760  "        "       "     1.00 760 

"     30.  "500"         "       "     1.50 750 

When  is  the  whole  amount  of  the  above  bill  due,  per 
average?  ^W5.  June  18. 

5.  My  expenditures  in  building  a  house,  in  the  year  1856, 
were  as  follows  :  Jan.  16,  $536.78  ;  Feb.  20,  $425.36  ;  March 
4,  $259.25  ;  April  24,  $786.36.     If  at  the  last  date  I  agi'oe  to 


i 


AVERAGE    ACCOUNTS, 


277 


sell  the  house  for  exactly  what  it  cost,  with  reference  to  in- 
terest on  the  money  expended,  and  take  the  purchaser's  note 
for  the  amount,  what  shall  be  the  face  of  the  note,  and 
what  its  date  ?  ^^^     (  Pace,  $2007.75. 

t  Date,  March  8,  1856. 
a  Thomas  Whiting, 
1859.  To  IsEAEL  Palmer,     Dr. 

Jan.    1.    To    60  bbls.  Flour,    @  $7.00 $420 

''    28.      ''     90  bu.     Wheat,*'     1.50 135 

Mar.  15.      "  300  bbls.  Flour,    ''     6.00 1800 

If  credit  of  3  months  be  given  to  each  item,  when  will 
the  above  account  become  due  ?  Ans.  May  30. 

Case  III. 

363.  "When  the  terms  of  credit  begin  at  different 
times,  and  the  account  has  both  a  debt  and  a  credit 
side. 

1.  Average  the  following  account: 

David  Ware. 


Dr. 

Or 

. 

185 

June 
tt 

Oct. 

3. 
1 

16 
20 

To  Mdse 

*'  Draft,  3  mo. 
"  Cash 

400 
800 
250 

00 
00 
00 

185 
July 
Aug. 
Sept. 

S. 

4  By  Mdse 

20  "  Cash 

20  "   "  

200 
150 

500 

00 
00 
00 

Dr. 


OPERATION. 


Cr. 


Focal 
date. 


Due. 

da. 

Items. 

Prod. 

Due. 

da. 

108 
61 
80 

Items. 

Prod. 

June  1 
Sept.  19 
Oct.  20 

141 

31 

0 

400 
800 
250 

56400 
24800 

July  4 
Aug.  20 
Sept.  20 

200 
150 
500 

21600 

9150 

15000 

- 

1450 
850 

81200 
45750 

850 

45750 

Balances. 

600 

35450 

35450  -^  600  =  59  da.,  average  term  of  interest. 
Oct.  20  —  59  da.  =  Aug.  22,  balance  due. 


\Miat  is  Case  III  ?    Explain  operation. 


278 


EQUATION    OF     I'AYMENTS, 


Analysis.  In  the  above  operation  we  have  written  the  dates, 
showing  when  the  items  become  due  on  either  side  of  the  account, 
adding  3  days'  grace  to  the  time  allowed  to  the  draft.  The  latest 
date,  Oct.  20,  is  assumed  as  the  focal  date  for  both  sides,  and  the  two 
columns  marked  da.  show  the  difference  in  days  between  each  date 
and  the  focal  date.  The  products  are  obtained  as  in  the  last  case,  and 
ft  balance  is  struck  between  the  items  charged  and  the  products. 
These  balances,  being  on  the  Dr.  side,  show  that  David  Ware,  on  the 
iay  of  the  focal  date,  Oct.  20,  owes  $600  with  interest  on  $35450  for 
1  day.  By  division,  this  interest  is  found  to  be  equal  to  the  interest 
on  $600  for  59  days.  The  balance,  $600,  therefore,  has  been  due  59 
days.  Reckoning  back  from  Oct.  12,  we  find  the  date  when  the  bal- 
ance fell  due,  Aug.  22. 

Rule.  I.  Find  the  time  when  each  item  of  the  account  is 
due  ;  and  write  the  dates,  in  tivo  columns,  on  the  sides  of  the 
account  to  which  they  respectively  helong. 

II.  Use  either  the  earliest  or  the  latest  of  these  dates  as  the 
focal  date  for  loth  sides,  and  find  the  prodiicts  as  in  the  last 
case, 

III.  Divide  the  halance  of  the  products  hy  the  balance  of 
the  account ;  the  quotient  will  be  the  interval  of  time,  which 
must  be  reckoned  from  the  focal  date  toward  the  other  dates 
when  both  bala^ices  are  on  the  same  side  of  the  account,  but 
FROM  the  other  dates  when  the  balances  are  on  opposite  sides 
of  the  account. 

2.  What  is  the  balance  of  the  following  account,  and  when 
is  it  due  ? 

JOHK  WiLSOK. 
Dr.  Or. 


1859. 

1 

1859.  1 

Jan. 

1 

To  Mdse. 

448 

00 

Jan. 

20 

Feb. 

4 

"   Cash. 

364 

00 

Feb. 

16 

It 

20 

€1           it 

232 

00 

tt 

25 

By  Am't  bro't  forward 

"   1  Carriage 

"   Cash 


560 
264 


900  00 


Ans. 


j  Balance,  $680. 
/  Due  March  13. 


3.  If  the  following  account  be  settled  by  giving  a  note, 
what  shall  be  the  face  of  the  note,  and  what  its  date  ? 


Qive  anal^'sis.    Bule. 


RATIO, 


279 


Isaac  Fostee. 


Br. 

Oi 

1858. 

1858. 

Jan. 

1 

To  Mdse. 

on  3  mo. 

145 

86 

May 

11 

By  Cash.... 

11 

00 

ti 

12 

U              ti 

..  5   «i 

37 

48 

July 

12 

«       « 

15 

00 

June 

3 

u           « 

«  3  " 

12 

25 

Oct. 

12 

**      * 

82 

00 

Aug. 

4 

((        «* 

M     O        <« 

66 

48 

j«o   i  $154.07,  face  of  note. 
^^'  \  Mar.  26, 1858,  date. 

EATIO. 

864.  Ratio  is  the  comparison  with  each  other  of  two 
numbers  of  the  same  kind.  It  is  of  two  kinds — arithmetical 
and  geometrical. 

365.  Aritlimetical  Katio  is  the  difference  of  the  two 
numbers. 

366.  Geometrical  Ratio  is  the  quotient  of  one  num- 
ber divided  by  the  other. 

367.  When  we  use  the  word  ratio  alone,  it  implies  geo- 
metrical ratio,  and  is  expressed  by  the  quotient  arising  from 
di^dding  one  number  by  the  other.  Thus,  the  ratio  of  4  to  8 
is  2,  of  10  to  5  is  J,  etc. 

368.  Katio  is  indicated  in  two  ways. 

1st.  By  placing  two  points  between  the  numbers  com- 
pared, writing  the  divisor  before  and  the  dividend  after  the 
points.  Thus,  the  ratio  of  5  to  7  is  written  5:7;  the  ratio 
of  9  to  4  is  written  9  :  4. 

2d.  In  the  form  of  a  fraction  ;  thus,  the  ratio  of  9  to  3 
is  I ;  the  ratio  of  4  to  6  is  f . 

369.  The  Terms  are  the  two  numbers  compared* 

370.  The  Antecedent  is  the  first  term. 

371.  The  Consequent  is  the  second  term. 

372.  No  comparison  of  two  numbers  can  be  fully  ex- 
plained but  by  instituting  another  comparison ;  thus,  the 

It  is  thongbt  best  to  omit  the  qneptione  at  the  bottom  of  the  pages  in  the  remain- 
ing part  of  this  work,  leaving  the  teacher  to  use  such  as  may  be  deemed  appropriate. 


280  RATIO. 

comparison  or  relation  of  4  to  8  cannot  be  fully  expressed 
by  2,  nor  of  8  to  4  by  |^.  If  the  question  were  asked,  what 
relation  4  bears  to  8,  or  8  to  4,  in  respect  to  magnitude,  the 
answer  2,  or  J,  would  not  be  complete  nor  correct.  But  if 
we  make  unity  the  standard  of  comparison,  and  use  it  as 
one  of  the  terms  in  illustrating  the  relation  of  the  two 
numbers,  and  say  that  the  ratio  or  relation  of  4  to  8  is  the 
same  as  1  to  2,  or  the  ratio  of  8  to  4  is  the  same  as  1  to  -J, 
unity  in  both  cases  being  the  standard  of  comparison,  then 
the  whole  meaning  is  conveyed. 

373.  A  Direct  Ratio  arises  from  dividing  the  conse- 
quent by  the  antecedent.   . 

374.  An  Inverse  or  Reciprocal  Ratio  is  obtained  by 
dividing  the  antecedent  by  the  consequent.  Thus,  the  direct 
ratio  of  5  to  15  is  -1^  =  3 ;  and  the  inverse  ratio  of  5  to  15 
is  ^j  =  h 

375.  A  Simple  Ratio  consists  of  a  single  couplet ;  as 
3:12. 

376.  A  Compound  Ratio  is  the  product  of  two  or  more 
simple  ratios.  Thus,  the  compound  ratio  formed  from  the 
simple  ratios  of  3  :  6  and  8:2is|-x|  =  3x8:6x2  = 

377.  In  comparing  numbers  with  each  other,  they  must 
be  of  the  same  kind,  and  of  the  same  denomination, 

378.  The  ratio  of  two  fractions  is  obtained  by  dividing 
the  second  by  the  first ;  or  by  reducing  them  to  a  common 
denominator,  when  they  are  to  each  other  as  their  numera- 
tors. Thus,  the  ratio  of  -^  :  |  is  |  -^  ^  =  f^  =  2,  which  is 
the  same  as  the  ratio  of  the  numerator  3  to  the  numerator 
6  of  the  equivalent  fractions  ^  and  ^. 

Since  the  antecedent  is  a  divisor  and  the  consequent  a  divi- 
dend, any  change  in  either  or  both  terms  will  be  governed  by 
the  general  principles  of  division,  (  87.)  We  have  only  to 
substitute  the  terms  antecedent,  consequent,  and  ratio,  for 
divisor,  dividend,  and  quotient,  and  these  princij^les  become 


EATIO.  281 

GENERAL    PRINCIPLES    OF    RATIO. 

Pkik.  I.  Multiplying  the  consequent  multiplies  the  ratio; 
dividing  the  consequent  divides  the  ratio. 

Pkij^.  II.  Multiplying  the  antecedent  divides  the  ratio ; 
dividing  the  antecedent  multiplies  the  ratio. 

Prik.  III.  Multiplying  or  dividing  loth  antecedent  and 
consequent  hy  the  same  number  does  not  alter  the  ratio. 

These  three  principles  may  be  embraced  in  one 

Gen"eeal  Law. 

A  change  in  the  consequeis't  produces  a  like  change  in 
the  ratio;  tut  a  change  in  the  antecedent  produces  an 
OPPOSITE  change  in  the  ratio. 

379.  Since  the  ratio  of  two  numbers  is  equal  to  the  con- 
sequent  divided  by  the  antecedent,  it  follows,  that 

1.  The  antecedent  is  equal  to  the  consequent  divided  by 
the  ratio  ;  and  that, 

2.  The  consequent  is  equal  to  the  antecedent  multiplied 
by  the  ratio. 

Examples  for  Practice. 

I.  What  part  of  9  is  3  ? 

I  =  -^  ;  or,  9  : 3  as  1 :  ^,  that  is,  9  has  the  same  ratio  to  3 
that  1  has  to  \. 

2.  What  part  of  20  is  5  ?  Ans.  \. 

3.  What  part  of  36  is  4  ?  Ans.  \. 

4.  What  part  of  7  is  49  ?  Ans.  7  times. 

5.  What  is  the  ratio  of  16  to  88  ?  Ans.  h\. 

6.  What  is  the  ratio  of  6  to  8|?  Ans.  |J. 

7.  What  is  the  ratio  of  %\  to  78  ?  Ans.  12. 

8.  What  is  the  ratio  of  16  to  m  ?  Ans.  4J. 

9.  What  is  the  ratio  of  J  to  |  ?  Ans.  \. 
10.  What  is  the  ratio  of  |  to  ^^  ?  Ans.  f . 

II.  What  is  the  ratio  of  3^  to  16|?  Ans.  5. 
12.  What  is  the  ratio  of  3  gal.  to  2  qt.  1  pt  ?     Ans.  ^. 


282  PROPORTION. 

13.  What  is  the  ratio  of  6.3s.  to  8s.  6d.?         Ans.  Iff. 

14.  What  is  the  ratio  of  5.6  to  .56  ?  Ans.  ■^. 

15.  What  is  the  ratio  of  19  Ihs.  5  oz.  8  pwts.  to  25  lbs. 
11  oz.  4  pwts.  ?  Ans.  IJ. 

16.  What  is  the  inverse  ratio  of  12  to  16?  Ans.  }. 

17.  What  is  the  inverse  ration  of  f  to  |  ?  Ans.  ^. 

18.  What  is  the  inverse  ratio  of  5J  to  17}?         Ans.  -J. 

19.  If  the  consequent  be  16  and  the  ratio  2f,  what  is  the 
antecedent?  Ans.  7. 

20.  If  the  antecedent  be  14.5  and  the  ratio  3,  what  is  the 
consequent  ?  Ans.  43.5. 

21.  If  the  consequent  be  J  and  the  ratio  J,  what  is  the 
antecedent?  Ans.  1|-. 

22.  If  the  antecedent  be  |  and  the  ratio  |,  what  is  the 
consequent?  Ans.  ^. 

PROPORTION. 

380.  Proportion  is  an  equaHty  of  ratios.  Thus,  the 
ratios  6  :  4  and  12  :  8,  each  being  equal  to  |,  form  a  proportion. 

381,  Proportion  is  indicated  in  two  ways  : 

1st.  By  a  double  colon  placed  between  the  two  ratios; 
thus,  2  :  5  : :  4  :  10. 

2d,  By  the  sign  of  equality  placed  between  the  two  ratios  ; 
thus,  2  :  5  =  4  :  10. 

383.  Since  each  ratio  consists  of  two  terms,  every  pro- 
portion must  consist  of  at  least  fotir  terms. 

383.  The  Extremes  are  the  first  and  fourth  terms. 

384.  The  Means  are  the  second  and  third  terms. 

385.  Three  numbers  may  be  in  proportion  when  the  first) 
is  to  the  second  as  the  second  is  to  the  third.  Thus,  the 
numbers  3,  9,  and  27  are  in  proportion  since  3  :  9  : :  9  :  27, 
the  ratio  of  each  couplet  being  3. 

In  such  a  proportion  the  second  term  is  said  to  be  a  mean 
proportional  between  the  other  two. 

386.  In  every  proportion  the  product  of  the  extremes  is 
equal  to  the  product  of  the  means.  Thus,  in  the  proportion 
3  :  5  : :  6  :  10  we  have  3  x  10  =  5  x  6. 


SIMPLE    PROPORTION-.  283 

387.  Four  numbers  that  are  proportional  in  the  direct 
order  are  proportional  by  inversion,  and  also  by  alternation, 
or  by  inverting  the  means.  Thus,  the  proportion  2:3:: 
6  :  9,  by  inversion  becomes  3  :  2  :  :  9  :  6,  and  by  alterna- 
tion 2  :  6  :  :  3  :  9. 

388.  From  the  preceding  principles  and  illustrations,  it 
follows  that,  any  three  terms  of  a  proportion  being  given, 
the  fourth  may  readily  be  found  by  the  following 

EuLE.  I.  Divide  the  product  of  the  extremes  ly  one  of  the 
means,  and  the  quotient  will  he  the  other  mean.     Or, 

II.  Divide  the  product  of  the  means  hy  07ie  of  the  extremes, 
and  the  quotient  will  be  the  other  extreme. 

Examples  for  Practice. 
Find  the  term  not  given  in  each  of  the  following  proportions : 


1.  48  :  20  : :  (  )  :  50. 

Ans.  120. 

2.  42  :  70  :  :  3  :  (  ). 

Ans,  5. 

3.  (  )  :  30  :  :  20  :  100. 

Ans,  6. 

4.  1  :  (  )  : :  7  :  84. 

Ans.  12. 

5.  48  yd.  :  (  )  : :  $67.25  :  $201.75. 

Ans.  144  yd. 

6.  3  lb.  12  oz.  :  (  )  :  :  $3.50  :  $10.50. 

Ans,  11  lb.  4  oz 

7.  (  )  :  $38.25  ::  8bu.  2pk.  :  76bu.2pk.  Ans.U.26. 

8.  4|-  :  38i  : :  (  )  :  76^. 

Ans.  8^. 

9.  (  )  :12::i:  If 

Ans.  7. 

10.  A:  (  )::i:f. 

Ans.  |. 

SIMPLE    PROPORTION. 

389.  Simple  Proportion  is  an  equality  of  two  simple 
ratios,  and  consists  of  four  terms,  any  three  of  which  being 
given,  the  fourth  may  readily  be  found. 

390.  Every  question  in  simple  proportion  involves  the 
principle  of  cause  and  effect. 

391.  Causes  may  be  regarded  as  action,  of  whatever 
kind,  the  producer,  the  consumer,  men,  animals,  time,  dis- 
tance, weight,  goods  bought  or  sold,  money  at  interest,  etc. 


284  PROPORTION. 

393.  Effects  may  be  regarded  as  whatever  is  accom- 
plished by  action  of  any  kind,  the  thing  produced  or  con- 
sumed, money  paid,  etc. 

393.  Causes  and  effects  are  of  two  kinds — simple  and 
compound. 

394.  A  Simple  Cause,  or  Effect,  contains  but  one  ele- 
ment ;  as  goods  purchased  or  sold,  and  the  money  paid  or 
received  for  them. 

395.  A  Compound  Cause,  or  Effect,  is  the  product  of 
two  or  more  elements  ;  as  men  at  work  taken  in  connection 
with  time,  and  the  result  produced  by  them  taken  in  con- 
nection with  dimensions,  length  and  breadth,  etc. 

396.  Causes  and  effects  that  admit  of  computation,  that 
is,  involve  the  idea  of  quantity ,  may  be  represented  by  num- 
bers, which  will  have  the  same  relation  to  each  other  as  the 
things  they  represent.  And  since  it  is  a  principle  of  philoso- 
phy that  like  causes  produ'ce  like  effects,  and  that  effects  are 
always  in  proportion  to  their  causes,  we  have  the  following 
proportions : 

1st  Cause  :  2d  Cause  : :  1st  Effect  :  2d  Effect. 
Or,  1st  Effect  :  2d  Effect  ; :  1st  Cause  :  2d  Cause  ; 
in  which  the  two  causes,  or  the  two  effects  forming  one  coup- 
let, must  be  like  numbers,  and  of  the  same  denomination. 

Considering  all  the  terms  of  the  proportion  as  abstract 
numbers,  we  may  say  that 

1st  Cause  :  Ist  Effect  : :  2d  Cause  :  2d  Effect ; 

which  will  produce  the  same  numerical  result. 

But  as  ratio  is  the  result  of  comparing  tAvo  numbers  or 
things  of  the  same  kind  (377),  the  first  form  is  regarded 
as  the  most  natural  and  philosophical. 

397.  Simple  causes  and  simple  effects  give  rise  to  simple 
ratios  ;  compound  causes  and  compound  effects  to  compound 
ratios. 

398.  1.  If  5  tons  of  coal  cost  $30,  what  will  3  tons  cost  ? 

The  required  term  will  be  denoted  by  a  (   ),  and  designated  "  blank." 


SIMPLE    PROPOHTION.  285 

STATEMENT.  ANALYSIS.     In  this 

tons.       tons.  $  $  example  an  effect  is 

5      :     3      :  :     30     :      (     )  required,  and  5  tons 

Ist  cause.  2d  cause.     Ist  effect.     2d  effect.  must  have  the  same 

OPERATION.  ratio  to  3  tons,  as  $30, 

5x(     )  =  3x30  the  cost  of  5  tons,  to 

3  ><  ^06  (blank)    dollars,  the 

(     )  =  rt —    =  ^^^>  ^^^-  cost  of  3  tons. 

Since  the  product 
of  the  extremes  is  equal  to  the  product  of  the  means  (373),  and  the 
product  of  the  means  divided  by  one  of  the  extremes  will  give  the 
other ;  (blank)  dollars  will  be  equal  to  the  product  of  3  x  30  divided  by 
5,  which  is  $18. 

2.  If  15  barrels  of  flour  cost  $90,  how  many  barrels  can 
be  bought  for  $30  ? 

Analysis.  In  this  ex- 
ample acaiLse  is  required, 
and  the  statement  may  be 
read  thus  :  If  15  barrels 
cost  $90,  how  many  or 
(blank)  barrels  will  cost 
$30  ?  The  product  of  the 
extremes,  30  x  15,  divided 
(     )  =  5  bar.,  Ans,  ^^  ^he  given  mean,  90, 

will  give  the  required  term,  5,  as  shown  in  the  operation. 

KuLE.  I.  Arrange  the  terms  in  the  statement  so  that  the 
causes  shall  compose  07ie  couplet,  and  the  effects  the  other, 
putting  (     )  in  the  place  of  the  required  term. 

II.  If  the  required  term  he  an  extreme,  divide  the  product 
of  the  means  hy  the  given  extreme ;  if  the  required  term  he  a 
mean,  divide  the  product  of  the  extremes  hy  the  given  mean. 

1.  If  the  terms  of  any  couplet  be  of  different  denominations,  they  must  be  reduced 
to  the  same  unit  value. 

2.  If  the  odd  term  be  a  compound  number,  it  must  be  reduced  to  its  lowest  imit. 

3.  If  the  divisor  and  dividend  contain  one  or  more  factors  common  to  both,  they 
should  be  canceled.  If  any  of  the  terms  of  a  proportion  contain  mixed  numbers, 
they  should  first  be  changed  to  improper  fractions,  or  the  fractional  part  to  a  decimal. 

4.  When  the  vertical  line  is  used,  the  divisor  and  the  required  term  are  written  on 
the  left,  and  the  terms  of  the  dividend  on  the  right. 


STATEMENT. 

bar. 

bar.                   $ 

% 

15 

•     (    )     ::    90 

:    30 

1st  cause. 

2d  cause.       1st  effect. 
OPERATION. 

2d  effect. 

00 

n 

(       ) 

u^ 

286  PROPORTION". 

399.  We  will  now  give  another  method  of  solving  ques- 
tions in  simple  proportion,  without  making  the  statement, 
and  which  may  be  used,  by  those  who  prefer  it,  to  the  one 
already  given.     We  will  term  it  the 

SECoiirD  Method. 

Every  question  which  properly  belongs  to  simple  propor- 
tion must  contain  four  numbers,  at  least  three  of  which 
must  be  given  (389).  Of  the  three  given  numbers,  one 
must  always  be  of  the  same  denomination  as  the  required 
number.  The  remaining  two  will  be  like  numbers,  and 
bear  the  same  relation  to  each  other  that  the  third  does  to 
the  required  number  ;  in  other  words,  the  ratio  of  the  third 
to  the  required  number  will  be  the  same  as  the  ratio  of  the 
other  two  numbers. 

Regarding  the  third  or  odd  term  as  the  antecedent  of  the 
second  couplet  of  a  proportion,  we  find  the  consequent  or  re- 
quired term  by  multiplying  the  antecedent  by  the  ratio  (379). 

By  comparing  the  two  like  numbers,  in  any  given  ques- 
tion, with  the  third,  we  may  readily  determine  whether  the 
answer,  or  required  term,  will  be  greater  or  less  than  the 
third  term ;  if  greater,  then  the  ratio  will  be  greater  than 
1,  and  the  two  like  numbers  may  be  arranged  in  the  form 
of  an  improper  fraction  as  a  multiplier ;  if  the  answer,  or 
required  term,  is  to  be  less  than  the  third  term,  then  the 
ratio  will  be  less  than  1,  and  the  two  like  numbers  may  bo 
arranged  in  the  form  of  a  proper  fraction,  as  a  multiplier. 

1.  If  4  cords  of  wood  cost  $12,  what  will  20  cords  cost? 
OPERATION.  Analysis.    It  will 

1^3  X  20       ^^^  be  readily  seen  in  this 

12  X  ^,  written  — ^ =  $60.  example,  that  4  cords 

and  20  cords  are  the 
like  terms,  and  that  $12  is  the  third  term,  and  of  the  same  denomina- 
tion as  the  answer  or  required  term. 

If  4  cords  cost  $12,  will  20  cords  cost  more,  or  less,  than  4  cords  7 
evidently  more  :  then  the  answer  or  required  term  will  be  greater 


SIMPLE    PBOPORTION-.  287 

than  fhe  third  term,  and  the  ratio  greater  than  1.  The  ratio  of  4 
cords  to  20  cords  is  ^^,  or  5 ;  hence  the  ratio  of  $12  to  the  answer 
must  be  5,  and  the  answer  will  be  -®^  or  5  times  $12,  which  is  $60. 

2.  If  12  yards  of  cloth  cost  $48,  what  will  4  yards  cost  ? 

OPERATION.  Analysis.    In  this  example  we 

48  X  ^  =  116,  Ans.  see  that  12  yards  and  4  yards  are 

the  like  terms  and  $48  the  third 
term,  and  of  the  same  denomination  as  the  required  answer. 

If  12  yards  cost  $48,  will  4  yards  cost  more  or  less  than  12  yards  ? 
less :  then  the  ratio  will  be  less  than  1,  and  the  multiplier  a  proper 
fraction.  The  ratio  of  12  yards  to  4  yards  is  3% ;  hence  the  ratio  of  $48 
to  the  answer  is  j\,  and  the  answer  will  be  j%  times  $48,  which  is  $16. 

Rule.  I.  With  the  two  given  numbers,  which  are  of  the 
same  name  or  kind,  form  a  ratio  greater  or  less  than  1, 
according  as  the  answer  is  to  be  greater  or  less  than  the 
third  given  number. 

11.  Multiply  the  third  number  by  this  ratio,  and  the  pro- 
duct will  be  the  required  number  or  answer. 

1.  Mixed  numbers  should  first  be  reduced  to  improper  fractions,  and  the  ratio  of 
the  fractions  found  according  to  (378). 

2.  Reductions  and  cancellation  may  be  applied  as  in  the  first  method. 

The  following  examples  may  be  solved  by  either  of  the 
foregoing  methods. 

Examples  foe  Practice. 

1.  If  48  cords  of  wood  cost  $120,  what  will  20  cords  cost  ? 

Ans.  $50. 

2.  If  6  bushels  of  com  cost  $4.75,  what  will  75  bushels 
cost  ?  Ans.  $59.37f 

3.  If  8  yards  of  cloth  cost  $3|^,  how  many  yards  can  be 
bought  for  $50  ?  Ans.  114f  yds. 

4.  If  12  horses  consume  42  bushels  of  oats  in  3  weeks,  how 
many  bushels  will  20  horses  consume  in  the  same  time  ? 

5.  If  7  pounds  of  sugar  cost  75  cents,  how  many  pounds 
can  be  bought  for  $9  ?  Ans,  84  lbs. 

6.  What  will  11  lb.  4  oz.  of  tea  cost,  if  3  lb.  12  oz.  cost 
13.50?  Ans.  $10.50. 


288  SIMPLE    PROPORTION. 

7.  If  a  staff  3  ft.  8  in.  long  cast  a  shadow  1  ft.  6  in.,  what 
is  the  height  of  a  steeple  that  casts  a  shadow  75  feet  at  the 
same  time?  Ans.  183  ft.  4  in. 

8.  At  12.75  for  14  pounds  of  sugar,  what  will  be  the  cost 
of  100  pounds  ?  .  Ans.  $19.6-4f. 

9.  How  many  bushels  of  wheat  can  be  bought  for  $51.06, 
if  12  bushels  can  be  bought  for  $13.32  ? 

10.  What  will  be  the  cost  of  28J  gallons  of  molasses,  if 
15  hogsheads  cost  $236.25  ?  A?is.  17.12^. 

11.  If  7  barrels  of  flour  are  sufficient  for  a  family  6  months, 
how  many  barrels  will  they  require  for  11  months  ? 

12.  At  the  rate  of  9  yards  for  £5  12s.,  how  many  yards  of 
cloth  can  be  bought  for  £44  16s.  ?  Ans.  72  yds. 

13.  An  insolvent  debtor  fails  for  $7560,  of  which  he  is 
able  to  pay  only  $3100 ;  how  much  will  A  receive,  whose 
claim  is  $756  ?  Ans.  $310. 

14.  If  2  pounds  of  sugar  cost  25  cents,  and  8  pounds  of 
sugar  are  worth  5  pounds  of  coffee,  what  will  100  pounds 
of  coffee  cost  ?  Atis.  $20. 

15.  If  the  moon  move  13°  10'  35"  in  1  day,  in  what  time 
will  it  perform  one  revolution  ? 

16.  If  8|  bushels  of  corn  cost  $4.20,  what  will  be  the  cost 
of  13^  bushels  at  the  same  rate  ?  Ajis.  $6.48. 

17.  If  If  yards  of  cotton  cloth  cost  6J  pence,  how  many 
yards  can  be  bought  for  £10  6s.  8d.  ?  A71S.  694|  yds. 

18.  If  12i  cwt.  of  iron  cost  $42^,  what  will  48|  cwt.  cost? 

19.  What  quantity  of  tobacco  can  be  bought  for  $317.23, 
if  8|  lbs.  cost  $1|?  Ans.  15  cwt.  22.7+  lbs. 

20.  If  15|  bushels  of  clover  seed  cost  $156J,  how  much 
can  be  bought  for  $95.75  ?  Ans.  9  bu.  2  pk.  2|  qt 

21.  If  I  of  a  baiTel  of  cider  cost  ^^,  how  much  will  -J  of 
a  barrel  cost  ?  Ans.  ^■^. 

22.  If  a  piece  of  land  of  a  certain  length,  and  4  rods  in 
breadth,  contain  |  of  an  acre,  how  much  would  there  bo  if 
it  were  11|  rods  wide?  Ans.  2  A.  28  rods. 

23.  If  13  cwt.  of  iron  cost  $42^,  what  will  12  cwt.  cost  ? 


SIMPLE  PKOPORTio:sr.  289 

24.  A  grocer  has  a  false  balance,  by  which  1  pound  will 
weigh  but  12  oz. ;  what  is  the  real  value  of  a  barrel  of  sugar 
that  he  sells  for  $28  ?  Ans.  $21. 

25.  A  butcher  in  selling  meat  sells  144-J-  oz.  for  a  pound  ; 
how  much  does  he  cheat  a  customer,  who  buys  of  him  to  the 
amount  of  $30  ?  Ans.  $2.46  + . 

26.  If  a  man  clear  $750  by  his  business  in  1  yr.  6  mo., 
how  much  would  he  gain  in  3  yr.  9  mo.  at  the  same  rate  ? 

27.  If  a  certain  business  yield  $350  net  profits  in  10  mo., 
in  what  time  would  the  same  business  yield  $1050  profits? 

28.  B  and  0  have  each  a  farm ;  B's  farm  is  worth  $25 
an  acre,  and  O's  $30 J  ;  if  in  trading  B  values  his  land  at  $28 
an  acre,  what  value  should  0  put  upon  his  ?   Ans.  $34.16. 

29.  If  I  borrow  $500,  and  keep  it  1  yr.  4  mo.,  for  how 
long  a  time  should  I  lend  $240  as  an  equivalent  for  the 
favor  ?  Ans.  2  yr.  9  mo.  10  da. 

COMPOUND   PKOPORTION. 

400.  Compound  Proportion  embraces  that  class  of 
questions  in  which  the  causes,  or  the  effects,  or  both,  are 
compound. 

The  required  term  may  be  a  cause,  or  a  single  element  of  a 
cause ;  or  it  may  be  an  effect,  or  a  single  element  of  an  effect. 

1.  If  16  horses  consume  128  bushels  of  oats  in  50  days, 
how  many  bushels  will  5  horses  consume  in  90  days  ? 

STATEMENT. 
let  cause.  2d  cause.       Ist  effect.       2d  effect. 

1   S     =     1    9^     ==    ^^«     =     (    ) 
Or,       16  X  50      :  5  X  90     : :    128     :     (    ) 

OPEKATiON.  Analysis.    In  tMs  ex- 

$  X  ^0^  X  ^t$^  ample  the  required  term 

^0~X~^0  ^  is  the  second  effect ;  and 

the  question  may  be  read. 
If  16  horses  in  50  days  consume  128  bushels  of  oats,  5  horses  in  90 
days  will  consume  how  many,  or  (blank)  bushels? 
These  questions  are  most  readily  performed  by  cancellation. 


290 


PROPORTION. 


2.  If  $480  gain  $84  interest  in  30  months,  what  sum  will 
gain  121  in  15  months  ? 


1st  canse. 

480 
30 


STATEMENT. 

2d  cause.      let  effect    2d  effect 


(480      .       j 

(    30     •      ( 


(   ) 
15 


OPEKATION. 


4$0^' 


X  n 


X  U 


=  $240,  Ans. 


84     :    21 

Analysis.  The  le. 
quired  term  in  this  ex- 
ample is  an  element  of 
the  second  clause  :  and 


the  question  may  be  read.  If  $480  in  30  months  gain  $84,  what  prin- 
cipal in  15  months  will  gain  $21  ? 

3.  If  7  men  dig  a  ditch  60  feet  long,  8  feet  wide,  and  6 
feet  deep,  in  12  days,  what  length  of  ditch  can  21  men  dig 
in  2|  days,  if  it  be  3  feet  wide  and  8  feet  deep  ? 


STATEMENT. 


Ul:     1 


21 

^ 


Or,    7  X  12  :  21  X  I 


reo 

8 
6 

60  X  8  X  6 


0 

3 

8 

(  )  X  3  X  8. 


^1  X   8   X 


OPERATION. 
$0»    X    ^    X 


t  x:t$  X 
Or, 


$ 

8 

t 

n 

n 

00» 

$ 

$ 

3 

0» 

(  ) 

(    )=  80  ft,  Ans. 


Analysis.    In 

^  ^  ^  ^        .         this  example  the 

X  $~x~$  ~  ^^  ^**  ^q^^^®^  *^"^  ^ 

the  length  of  the 
ditch,  and  is  an  element  of  the 
second  effect.  The  question, 
as  stated,  will  read  thus :  if  7 
men,  in  13  days,  dig  a  ditch  60 
feet  long,  8  feet  wide,  and  6 
feet  deep,  21  men,  in  2f  days, 
will  dig  a  ditch  how  many,  or 
(blank)  feet  long,  3  feet  wide, 
and  8  feet  deep? 


Rule.  I.  Of  the  given  terms,  select  tJiose  which  constitute 
the  causes,  and  those  which  constitute  the  effects,  a7id  arrange 
them  in  couplets,  putting  (    )  in  place  of  the  required  term. 


COMPOUKD    PROPORTION-.  291 

11.  Then,  if  the  Uanh  term  (  )  occur  in  either  of  the  ex- 
tremes,  make  the  product  of  the  means  a  dividend,  and  the 
product  of  the  extremes  a  divisor  ;  hut  if  the  blank  term  occur 
in  either  mean,  make  the  product  of  the  extremes  a  dividend, 
and  the  product  of  the  rneans  a  divisor, 

1.  The  causes  must  be  exactly  alike  in  the  number  and  Hnd  of  their  terms ;  tha 
same  is  trae  of  the  effects. 

2.  The  same  preparation  of  the  terms  by  reduction  is  to  be  observed  as  in  simple 
proportion. 

401.  We  will  now  solve  an  example  according  to  the 
Second  Method  given  in  Simple  Proportion. 

1.  If  18  men  can  build  42  rods  of  wall  in  16  days,  how 
many  men  can  build  28  rods  in  8  days  ? 

Analysis.   In  this  example 

OPERATION.  n   xi       * 

all  the  terms  appear  m  coup- 

X$^  X  X  =:  24  men.       ^^*^'  except  one,  which  is  18 

^'^  $  men,  and  that  is  of  the  same 

kind  as  the  required  answer. 

Since  compound  proportion  is  made  up  of  two  or  more  simple  pro- 
portions, if  this  third  or  odd  term  be  multiplied  by  the  compound  ratio, 
or  by  the  simple  ratio  of  each  couplet  successively,  the  product  will 
be  the  required  term. 

By  comparing  the  terms  of  each  couplet  with  the  third  term  we  may 
readily  determine  whether  the  answer,  or  term  sought,  will  be  greater 
or  less  than  the  third  term ;  if  greater,  then  the  ratio  will  be  greater 
than  1,  and  the  multiplier  an  improper  fraction  ;  if  less,  the  ratio  will 
be  less  than  1,  and  the  multiplier  a  proper  fraction. 

First  we  will  compare  the  terms  composing  the  first  couplet,  43 
rods  and  28  rods,  with  the  third  term,  18  men.  If  42  rods  require 
\8  men,  how  many  men  will  28  rods  require?  less  men;  hence  the 
.•atio  is  less  than  1,  and  the  multiplier  a  proper  fraction,  f| ;  next,  if 
16  days  require  18  men,  how  many  men  will  8  days  require  ?  more 
men ;  hence  the  ratio  is  greater  than  1,  and  the  multiplier  an  improper 
fraction,  Y-.  Regarding  the  third  term  as  the  antecedent  of  a  couplet, 
the  consequent  being  the  term  sought,  if  we  multiply  this  third  term 
by  the  simple  ratios,  or  by  their  product,  we  shall  have  the  required 
term  or  answer,  thus :  18  x  |-|  x  -^g^  =  24,  as  shown  in  the  operation. 

2.  5  compositors,  in  16  days,  of  14  hours  each,  can  com- 
pose 20  sheets  of  24  pages  in  each  sheet,  50  lines  in  a  page. 


16 

u 

^ 

t 

X^ 

n 

0 

u 

102 

^0 

00 

m 

00 

293  COMPOUND    PROPORTION. 

and  40  letters  in  a  line ;  in  how  many  days,  of  7  hours  each, 
will  10  compositors  compose  a  volume  to  be  printed  in  the 
same  letter,  containing  40  sheets,  16  pages  in  a  sheet,  60 
lines  in  a  page,  and  50  letters  in  a  line  ?      Ans.  32  days. 

OPERATION. 
Days.  ^Comp.  Hours.   Sheets.  Pages.  Lines.  Letters. 

16xAxJ^xi*xifxttxtt  =  32  days. 

BY  CANCELLATION.  ANALYSIS.    The  required  term  or  an- 

swer is  to  be  in  days';  and  we  see  that 
all  the  terms  appear  in  pairs  or  couplets, 
except  the  16  days,  which  is  of  the  same 
kind  as  the  answer  sought. 

We  will  proceed  to  compare  the  terms 
of  each  couplet  with  the  16  days.  First, 
if  5  compositors  require  16  days,  how 
many  days  will  10  compositors  require? 
less  days;  hence  the  multiplier  is  the 
32  days,  Ans,  proper  fraction  y^^,  and  we  have  16  x  ^^, 

Next,  if  14  hours  a  day  require  16  days, 
how  many  days  will  7  hours  a  day  require  ?  more  days ;  hence  the 
multiplier  is  the  improper  fraction  \S  and  we  have  16  x  ^^^  x  \\ 
Next,  if  20  sheets  require  16  days,  how  many  days  will  40  sheets  re- 
quire ?  more  days ;  hence  the  multiplier  is  the  improper  fraction  f^, 
and  we  have  16  x  ^^  x  if^  x  |§.  Pursuing  the  same  method  with  the 
other  couplets,  we  obtain  the  result  as  shown  in  the  operation. 

Rule.  L  Of  the  terms  composing  each  couplet  form  a 
ratio  greater  or  less  than  1,  in  the  same  manner  as  if  the 
answer  depended  on  those  two  and  the  third  or  odd  term, 

II.  Multiply  the  third  or  odd  term  hy  these  ratios  succes- 
sively,  and  the  product  will  be  the  ayiswer  sought. 

By  the  odd  term  is  meant  the  one  that  is  of  the  same  kind  as  the  answer. 

The  following  examples  may  be  solved  by  eitlier  of  tho 
given  methods  : 

Examples  for  Practice. 
1.  If  IG  horses  consume  128  bushels  of  oats  in  50  days, 
how  many  bushels  will  5  horses  consume  in  1)0  days  ? 


COMPOUKD    PEOPORTIOX.  293 

2.  If  a  man  travel  120  miles  in  3  days  when  the  days  are 
12  hours  long,  in  how  many  days  of  10  hours  each  will  he 
require  to  travel  360  miles  ?  Ans.  lOf  days. 

3.  If  6  laborers  dig  a  ditch  34  yards  long  in  10  days,  how 
many  yards  can  20  laborers  dig  in  15  days  ?  Ans.  170  yds. 

4.  If  450  tiles,  each  12  inches  square,  will  pave  a  cellar, 
how  many  tiles  that  are  9  inches  long  and  8  inches  wide  will 
pave  the  same  ?  A7is.  900. 

5.  If  it  require  1200  yards  of  cloth  f  wide  to  clothe  500 
men,  how  many  yards  which  is  -J  wide  will  it  take  to  clothe 
960  men  ?  Ans,  3291^  yds. 

6.  If  8  men  will  mow  36  acres  of  grass  in  9  days,  of  9 
hours  each  day,  how  many  men  will  be  required  to  mow  48 
acres  in  12  days,  working  12  hours  each  day?    Ans,  6  men. 

7.  If  4  men,  in  2J  days,  mow  6-|  acres  of  grass  by  work- 
ing 8i  hours  a  day,  how  many  acres  will  15  men  mow  in  3| 
days  by  working  9  hours  a  day  ?  Ans.  40|^  acres. 

8.  If,  by  traveling  6  hours  a  day  at  the  rate  of  4|-  miles 
an  hour,  a  man  perform  a  journey  of  540  miles  in  20  days,  in 
how  many  days,  traveling  9  hours  a  day  at  the  rate  of  4|- 
miles  an  hour,  will  he  travel  600  miles?      Ans.  14f  days. 

9.  If  2J  yards  of  cloth  If  yards  wide  cost  I3.37J,  what 
cost  36 J  yards,  1^  yards  wide  ?  Ans.  $52.79+. 

10.  If  5  men  reap  52.2  acres  in  6  days,  how  many  men 
will  reap  417.6  acres  in  12  days?  Ans,  20  men. 

11.  If  6  men  dig  a  cellar  22.5  feet  long,  17.3  feet  wide,  and 
10.25  feet  deep,  in  2.5  days,  of  12.3  hours,  in  how  many  days, 
of  8.2  hours,  will  9  men  take  to  dig  another,  measuring  45 
feet  long,  34.6  wide,  and  12.3  deep?  Ans.  12  days. 

12.  If  54  men  can  build  a  fort  in  24J  days,  working  12^ 
hours  each  day,  in  how  many  days  will  75  men  do  the  same, 
when  they  work  but  10|-  hours  each  day?     Ans.  21  days. 

13.  If  24  men  dig  a  trench  33J  yards  long,  5f  wide,  and 
3^  deep,  in  189  days,  working  14  hours  each  day,  how  many 
hours  per  day  must  217  men  work,  to  dig  a  trench  23J-  yards 
long,  3f  wide,  and  2^  deep,  in  5^  days  ?     Ans.  16  hours. 


294  PAETNEESHIP. 

PAETNEESHIP. 

402.  Partnership  is  a  relation  established  between  two 
or  more  persons  in  trade,  by  which  they  agree  to  sliare  the 
profits  and  losses  of  business. 

403.  The  Partners  are  the  individuals  thus  associated. 

404.  Capital,  or  Stock,  is  the  money  or  property  in- 
vested in  trade. 

405.  A  Dividend  is  the  profit  to  be  divided. 

406.  An  Assessment  is  a  tax  to  meet  losses  sustained. 

Case  I. 

207.  To  find  each  partner's  share  of  the  profit  or 
loss,  when  their  capital  is  employed  for  equal  pe- 
riods of  time. 

1.  A  and  B  engage  in  trade ;  A  furnishes  $300,  and  B 
$400  of  the  capital ;  they  gain  $182 ;  what  is  each  one's 
share  of  the  profit  ? 

OPERATION.  Analysis.  Since 

$300  tlie  whole   capital 

^/^{^^  employed  is  $300 

5r;r7-  +  $400  =  $700,   it 

$700,  whole  Btock.  .^  ^^^^^^  ^^^^^ 

^%  —  f  J  A's  share  of  thestock.  furnishes    \l%  =  f 

4^^  =  -f,  B's  share  of  the  stock.  of  the  capital,  and 

$182  Xf  =    $78,  A'8  share  of  the  gain.  B  |g§  =  |  of  the 

$1 82  X  4  =  $104,  B's  share  of  the  gain.  capital.    And  since 

each  man's  share  of 
the  profit  or  loss  will  have  the  same  ratio  to  the  whole  profit  or  loss 
that  his  share  of  the  stock  has  to  the  whole  stock,  A  will  have  f  of 
the  entire  profit,  and  B  |,  as  shown  in  the  operation. 

We  may  also  regard  the  whole  capital  as  the  first  catise, 
and  each  man's  share  of  the  capital  as  the  second  cause ,  the 
whole  profit  or  loss  as  the  first  effect,  and  each  man's  share 
of  the  profit  or  loss  as  the  second  effect,  and  solve  by  propor- 
tion thus  ; 


PAETNEKSHIP. 


let  cause. 

3d  cause. 

Ist  effect 

ad  effect. 

$700    : 

$300 

:  :    $182 

:    (    ) 

$700    : 

$400 

:  :    $182 

:    (    ) 

»00 

$003 

tm 

**00« 

(     ) 

m^ 

f  ) 

ig^as 

(     )  =  $78,  A'8  profit  (     )  =  $104,  B'B  profit 

Rule.  Multiply  the  whole  profit  or  loss  by  the  ratio  of  the 
whole  capital  to  each  man's  share  of  the  capital.    Or, 

The  tuhole  capital  is  to  each  man's  share  of  the  capital  as  the 
whole  profit  or  loss  is  to  each  man's  share  of  the  profit  or  loss, 

2.  Three  men  trade  in  company ;  A  furnishes  $8000,  B 
$12000,  and  0  $20000  of  the  capital ;  their  gain  is  $1680 ; 
what  is  each  man's  share  ? 

Ans.  A's  $336  ;  B's  $504 ;  C's  $840. 

3.  Three  persons  purchased  a  house  for  $2800,  of  which 
A  paid  $1200,  B  $1000,  and  C  $600;  they  rented  it  for  $224 
a  year ;  how  much  of  the  rent  should  each  receive  ? 

4.  A  man  failed  in  business  for  $20000,  and  his  available 
means  amounted  to  only  $13654  ;  how  much  wiU  two  of  his 
creditors  respectively  receive,  to  one  of  whom  he  owes  $3060, 
and  to  the  other  $1530  ?         Ans,  $2089.062  ;  $1044.531. 

5.  Four  men  hired  a  coach  for  $13,  to  convey  them  to 
their  respective  homes,  which  were  at  distances  from  the 
place  of  starting  as  follows  :  A's  16  miles,  B's  24  miles,  C's 
28  miles,  and  D's  36  miles  ;  what  ought  each  to  pay  ? 

J        j  A  $2.     C  $3.50. 
^'''-    (B$3.     D$4.50. 

6.  A  captain,  mate,  and  12  sailors  took  a  prize  of  $2240, 
of  which  the  captain  took  14  shares,  the  mate  6  shares,  and 
each  sailor  1  share  ;  what  did  each  receive  ? 

7.  A  cargo  of  com,  valued  at  $3475.60,  was  entirely  lost; 
I  of  it  belonged  to  A,  J  of  it  to  B,  and  the  remainder  to  C  ; 
how  much  was  the  loss  of  each,  there  being  an  insurance  of 
$2512  ?     A71S,  $120.45,  A's.    $240.90,  B's.     $002.25,  C's. 


296  PARTNERSHIP- 

8.  Three  persons  engaged  in  the  lumber  trade ;  two  of  the 
persons  furnished  the  capital,  and  the  third  managed  the 
business ;  they  gained  $2571.24,  of  which  0  received  $6  as 
often  as  D  14,  and  E  had  ^  as  much  as  the  other  two  for 
taking  care  of  the  business  ;  how  much  was  each  one's  share 
of  the  gain?    Ans.  $12So.62,C's.  $857.08,  D's.  $428.54,  E's. 

9.  Four  persons  engage  in  the  coal  trade ;  D  puts  in 
$3042  capital ;  they  gain  $7500,  of  which  A  takes  $2000,  B 
$2800.75,  and  C  $1685.25  ;  how  much  capital  did  A,  B,  and 
C  put  in,  and  how  much  is  D's  share  of  the  gain  ? 

.^    (A,  $6000.         C,  $5055.75. 
'   (  B,  $8402.25.    D's  gain,  $1014. 

Case  IL 

408,  To  find  each  partner's  share  of  the  profit 
or  loss  when  their  capital  is  employed  for  U7iequal 
periods  of  time. 

It  is  evident  that  the  respective  shares  of  profit  and  loss 
will  depend  upon  two  conditions,  viz. :  the  amount  of  cajntal 
invested  by  each,  and  the  time  it  is  employed. 

1.  Two  persons  form  a  partnership ;  A  puts  in  $450  for 
7  months,  and  B  $300  for  9  months  ;  they  lose  $156  ;  how 
much  is  each  man's  share  of  the  loss  ? 

OPERATION.  Analysis.    The 

$450  X  7  =  $3150,  A'8  capital  for  1  mo.  use  of  $450  capital 

$300  X  9  =  $2700,  B'B      "      "      «  for  7  months  is  the 

— same  as  the  use  ot 

$5850,  entire  ' 7    tinges    $450,   or 

^44  =  tV,  A'b  Bhare  of  the  entire  capital  ^^^^^  ^^^  1  "^o"*^  *> 

MM  =  A    B'B     "       "         "  -  ^^*^  ^^  ^^^  ^«"  ^ 

IfT?       7?'      Ar^i  months  is  the  same 

$156  X  A  =  ^84,  A'sloBB.  astheuseof 9times 

$156  X  A  =  $72,  B'B    "  1300,  or  $2700  for 

1  month.  The  en- 
tire capital  for  1  month  is  equivalent  to  $3150  +  $3700  =  $5850.  If 
the  loss,  $156,  be  divided  between  the  two  i)artnerfl,  accordinp:  to 
Case  I,  the  results  will  be  the  loss  of  each  as  shown  in  the  operation. 


PARTNERSniP.  297 

Examples  of  this  kind  may  also  be  solved  by  proportion  as  in  Case  I, 
tbe  catises  being  compounded  of  capital  and  tvne  ;  thus, 

$5850  :  $3150  :  :  $156  :  (  ) 
$5850  :  $2700  :  :  $156  :  (  ) 


(     ) 


$1007  $$$0 

m^^  (  ) 


M00^ 


(      )  =  $84,  A's  loss.  {      )  ==  ^'^^^  B's  loss. 

Rule.  Multiply  each  man^s  capital  hy  the  time  it  is  em- 
ployed in  trade,  and  add  the  products.  Then  multiply  the 
entire  profit  or  loss  hy  the  ratio  of  the  sum  of  the  products  to 
each  product,  and  the  results  will  he  the  respective  shares  of 
profit  or  loss  of  each  partner.    Or, 

Multiply  each  man^s  capital  hy  the  time  it  is  employed  in 
trade,  and  regard  each  product  as  his  capital,  and  the  sum 
of  the  products  as  the  entire  capital,  and  solve  hy  proportion, 
as  in  Case  I. 

Examples  for  Practice. 

2.  Three  persons  traded  together ;  B  put  in  $250  for  6 
months,  C  $275  for  8  months,  and  D  $450  for  4  months ; 
they  gained  $825  ;  what  was  each  man's  shai'e  of  the  gain  ? 

3.  Two  merchants  formed  a  partnership  for  18  months. 
A  at  first  put  in  $1000,  and  at  the  end  of  8  months  he  put 
in  $600  more ;  B  at  first  put  in  $1500,  but  at  the  end  of  4 
months  he  drew  out  $300 ;  at  the  expiration  of  the  time 
they  found  that  they  had  gained  $1394.64  ;  what  was  each 
man's  share  of  the  gain?    Ans.  A's  $715.20  ;  B's  $679.44. 

4  Three  men  took  a  field  of  grain  to  harvest  and  thresh 
for  J-  of  the  crop  ;  A  furnished  4  hands  5  days,  B  3  hands 
6  days,  and  0  6  hands  4  days ;  the  whole  crop  amounted  to 
372  bushels  ;  what  was  each  man's  share  ? 

5.  William  Gallup  began  trade  January  1,  1856,  with  a 
capital  of  $3000,  and,  succeeding  in  business,  took  in  M.  H. 
Decker  as  a  partner  on  the  first  day  of  March  following,  with 
a  capital  of  $2000 ;  four  months  after  they  admitted  J.  New- 

13* 


298  ANALYSIS. 

man  as  third  partner,  who  put  in  $1800  capital ;  they  con- 
tinued their  partnership  until  April  1,  1858,  when  they 
found  that  $4388.80  had  been  gained  since  Jan.  1,  1856 ; 
what  was  each  one's  share  ?  C  $2106,  Gallup's. 

Ans.  ^  $1300,  Decker's. 

L  $  982.80,  Newman's. 

6.  Two  persons  engaged  in  partnership  with  a  capital  of 
$5600  ;  A'?  capital  was  in  trade  8  months,  and  his  share  of 
the  profits  was  $560 ;  B's  capital  was  in  10  months,  and  his 
share  of  the  profits  was  $800  ;  what  amount  of  capital  had 
each  in  the  firm?  Ans,  A,  $2613.33^  ;  B,  $2986.66f. 

7.  A,  B,  and  C,  engaged  in  trade  with  $1930  capital ;  A's 
money  was  in  3  months,  B's  5,  and  C's  7;  they  gained  $117, 
which  was  so  divided  that  ^  of  A's  share  was  equal  to  J  of 
B's  and  to  }  of  C's  ;  how  much  did  each  put  in,  and  what 
did  each  gain  ?  f  A,  $700  capital ;  $26  gain. 

Ans,  \  B,  $630      "         $39     " 
L  C,  $600      "         $52     " 

ANALYSIS. 

409.  Analysis,  in  arithmetic,  is  the  process  of  solving 
problems  independently  of  set  rules,  by  tracing  the  relations 
of  the  given  numbers  and  the  reasons  of  the  separate  steps 
of  the  operation  according  to  the  special  conditions  of  each 
question. 

410,  In  solving  questions  by  analysis,  we  generally  reason 
from  the  given  number  to  unity,  or  1,  and  then  from  unity, 
or  1,  to  the  required  number, 

411^  United  States  money  is  reckoned  in  dollars,  dimes, 
cents,  and  mills  ( 180 ),  one  dollar  being  uniformly  valued 
in  all  the  States  at  100  cents. 

At  the  time  of  the  adoption  of  onr  decimal  currency  by  Conprresp,  in  1786,  the 
colonial  currency,  or  billft  of  credit,  ispued  by  the  coloniee,  had  depreciated  in  value; 
and  this  depreciation,  being  unequal  in  the  different  colonies,  gave  riee  to  the  dif- 
ferent values  of  the  State  currencies. 

In  many  of  the  States  it  was  customary  to  pive  the  retail  price  of  articles  in  Blii)« 
Ungs  and  pence  and  the  cost  of  the  whole  in  dollars  and  cents. 


ANALYSIS.  299 

This  nsage  has  become  nearly  If  not  quite  obsolete  all  over  the  country ;  but  as 
the  matter  has  an  historical  interest,  it  is  retained  in  this  new  edition,  to  avoid 
derangement  with  previous  editions,  and  the  examples  will  also  afford  a  pleasant 
and  profitable  exercise  for  the  pupil. 

413.   In  New  England,  Indiana,   ) 
Illinois,  Missouri,  Virginia,  Kentucky,  )•  $1  =  6s.  =  72d. 
Tennessee,  Mississippi,  Texas ) 

New  York,  Ohio,  Michigan $1  =  8s.  =  96d. 

New  Jersey,   Pennsylvania,  Dela-  I  $i  __  7g  ga  _  qqa 
ware,  Maryland ) 

South  Carolina,  Georgia )   II  =  4s.  8d.  =  56d. 

The  Dominion  of  Canada )   |1  =  5s.  =  60d. 


Examples  for  Practice. 

1.  What  will  be  the  cost  of  42  bushels  of  oats,  at  3  shil' 
lings  per  bushel.  New  England  currency  ? 

OPERATION.  Analysis.    Since 

|4^^  1  bushel  costs  3  sliil- 

3  lings,  42  bushels  will 

x^u  -:-  u  ==  ^^x.     yji,  cost  42  times  3s.,  or 

$21,  Ans.  42  X  3  =  126s. ;  and 

as  6s.  make  1  dollar 
New  England  currency,  there  are  as  many  dollars  in  126s.  as  6  is  con- 
tained times  in  126,  or  $21. 

2.  What  will  180  bushels  of  wheat  cost  at  9s.  4d.  per 
bushel,  Pennsylvania  currency  ? 

OPERATION.  Analysis.      Multi. 

Or, 
00 


112  $ 


$224  ^^ 


plying  the  number  of 
28  bushels  by  the  price, 

o  and  dividing    by  the 

value  of  1  dollar  re- 

$224,  Ans,  duced    to    pence,   we 

have  $224.  Or,  when 
the  pence  in  the  ^ven  price  is  an  aliquot  part  of  a  shilling,  the  price 
may  be  reduced  to  an  improper  fraction  for  a  multiplier,  thus  :  9s.  4d. 
=  9|s.  =  -\^-s.,  the  multiplier.  The  value  of  the  dollar  being  7s.  6d.  = 
7|s.  =  ^',  we  divide  by  -i/,  as  in  the  operation. 

3.  What  will  be  the  cost  of  3  hhd.  of  molasses,  at  Is.  3d, 
per  quart,  Georgia  currency  ? 


800 


2 


$X 


BOO  AN-ALYSIS. 

OPEBATION. 

3  Analysis.    In  this  example  we  first 

0^9  reduce  3  lihd.  to  quarts,  by  multiplying 

^  by  63  and  4,  and  then  multiply  by  the 

^  f,  price,  either  reduced  to  pence  or  to  an 

improper  fraction,  and  divide  by  the 

405.00  value  of  1  dollar  reduced  to  the  same 

^202  50  denomination  as  the  price. 

4.  Sold  9  firkins  of  butter,  each  containing  56  lb.,  at  Is.  6d. 
per  pound,  and  received  in  payment  carpeting  at  6s.  9d. 
per  yard  ;  how  many  yards  of  carpeting  would  pay  for  the 
butter  ? 

OPERATION.  Analysis.    The  operation  in  this  is  similar 

0  to  the  preceding  examples,  except  that  we  di- 

gg  vide  the  cost  of  the  butter  by  the  price  of  a 

^  rt  3  unit  of  the  articl  e  received  in  payment,  reduced 

.  to  the  same  denominational  unit  as  the  price 

112  yd.  of  a  unit  of  the  article  sold.    The  result  will 
be  the  same  in  whatever  currency. 

5.  What  will  3  casks  of  rice  cost,  each  weighing  126 
pounds,  at  4d.  per  pound,  South  Carolina  currency  ? 

Ans,  127. 

6.  How  many  pounds  of  tea,  at  7s.  per  pound,  must  bo 
given  for  28  lb.  of  butter,  at  Is.  7d.  per  pound  ?    A7is.  6^. 

7.  Bought  2  casks  of  Catawba  wine,  each  containing  72 
gallons,  for  $648,  and  sold  it  at  the  rate  of  10s.  6d.  per  quart, 
Ohio  currency  ;  how  much  was  my  whole  gain  ?    Ans.  $108. 

8.  What  will  it  cost  to  keep  2  horses  3  wk.  if  it  cost  to 
keep  1  horse  1  day,  2s.  6d.  Canada  currency  ? 

9.  How  many  days'  work,  at  6s.  3d.  per  day,  must  be  given 
for  20  bushels  of  apples  at  3s.  per  bushel  ?  A7is.  9|. 

10.  Bought  160  lb.  of  dried  fruit,  at  Is.  6d.  a  pound,  in 
New  York,  and  sold  it  for  2s.  a  pound  in  Pliiladelphia ; 
what  was  my  whole  gain  ?  Ans.  $!l2.66-|. 

11.  A  merchant  exchanged  43|  yards  of  cloth,  worth 
lOs.  6d.  per  yard,  for  other  cloth  worth  8s.  3d.  per  yard ; 
how  many  yards  did  he  receive  ?  Ans.  55^^. 


ANALYSIS.  301 

12.  What  will  be  the  cost  of  300  bushels  of  wheat  at  9s.  4d. 
per  bushel,  Michigan  currency  ?  Ans.  1350. 

13.  If  f  of  4  of  a  ton  of  coal  cost  $2f ,  how  much  will  4  of 
6  tons  cost  ? 

OPERATION. 

^  I  1^2  Analysis.    Since  |  of  f  =  ^  of  a  ton  costs 

i:^  U04  $2|  =  $-V-,  1  ton  wiU  cost  28  times  ^  of  $J^, 

^     *03  or  $-i/xf|;  and  f  of  6  tons  =  V"  toi^»  will 

— -! cost  -^/  times  f|  of  $J^  =  $16. 

$16,  Ans, 

14.  If  8  men  can  build  a  wall  20  ft.  long,  6  ft.  high,  and 
4  ft.  thick,  in  12  days,  working  10  hours  a  day,  in  how  many 
days  can  24  men  build  a  wall  200  ft.  long,  8  ft.  high,  and  6  ft. 
thick,  working  8  hours  a  day  ? 


OPERATION. 

u 

X 

X 

10 

^0010 

X 

1 

^       ^0 

^  X  ^  =  100  da. 

Analysis.  Since  8  men  require  13  days  of  10  hours  each  to  build 
the  wall,  1  man  would  require  8  times  12  days  of  10  hours  each,  and 
10  times  (12  x  8)  days  of  1  hour  each.  To  build  a  wall  1  ft.  long  would 
require  ^^  as  much  time  as  to  build  a  wall  20  ft.  long  ;  to  build  a  wall 
1  ft.  high  would  require  ^  as  much  time  as  to  build  a  wall  6  ft.  high  ; 
to  build  a  wall  1  ft.  thick,  \  as  much  time  as  to  build  a  wall  4  ft.  thick. 
Now,  24  men  could  build  this  wall  in  ^  as  many  days,  by  working 
1  hour  a  day,  as  1  man  could  build  it,  and  in  J  as  many  days  by  work- 
ing 8  hours  a  day,  as  by  working  1  hour  a  day ;  but  to  build  a  wall 
200  ft.  long  would  require  200  times  as  many  days  as  to  build  a  wall 
1  ft.  long ;  to  build  a  wall  8  ft.  high  would  require  8  times  as  many 
days  as  to  build  a  wall  1  ft.  high  ;  and  to  build  a  wall  6  ft.  thick  would 
require  6  times  as  many  days  as  to  build  a  wall  1  ft.  thick. 

15.  If  2  pounds  of  tea  are  worth  11  pounds  of  coifee,  and 
3  pounds  of  coffee  are  worth  5  pounds  of  sugar,  and  18 
pounds  of  sugar  are  worth  21  pounds  of  rice,  how  many 
pounds  of  rice  can  be  purchased  with  12  pounds  of  tea  ? 


302  ANALYSIS. 

OPERATION.  Analysts.    Since  18  lb.  of  sugar 

^1 '  are  equal  in  value  to  21  lb.  of  rice, 

5  1  lb.  of  sugar  is  equal  to  y^^  of  21  lb. 

-J  ^  of  rice,  or  ^  =  -g  lb.  of  rice,  and  5  lb. 

of  sugar  are  equal  to  5  times  ^  lb. 

^^  of  rice,  or  5^  lb. ;  if  3  lb.  of  coffee 

3  I  385  are  equal  to  5  lb.  of  sugar,  or  \^  lb. 

.      "ToQj.  iv.  ®^  ^^^^>  1  ^^-  ^^  coffee  is  equal  to  ^ 

jlnS,    l/i»-^  ID.  ^^    8^5   j^    ^^  ^^^  ^j.   35   j^^  ^^  jj   jj^ 

of  coffee  are  equal  to  11  times  ff  lb.  of  rice,  or  -\«/  lb.;  if  3  lb.  of  tea 
are  equal  to  11  lb.  of  coffee,  or  ^^-  lb.  of  rice,  1  lb.  of  tea  is  equal  to  ^ 
of  i*y\5-  lb.  of  rice,  or  ^%^  lb.,  and  12  lb.  of  tea  are  equal  to  13  times  ^^ 
lb.  of  rice,  or  iJ|^  lb.  =  128^  lb. 

16.  If  16  horses  consume  128  bushels  of  oats  in  50  days, 
how  many  bushels  will  5  horses  consume  in  90  days  ?  Ans.  72. 

17.  If  $10J  will  buy  4f  cords  of  wood,  how  many  cords 
can  be  bought  for  $24 J  ?  Ans,  11. 

18.  Gave  52  barrels  of  potatoes,  each  containing  3  bushels, 
worth  33^  cents  a  bushel,  for  65  yards  of  cloth  ;  how  much 
was  the  cloth  worth  per  yard  ?  Ans,  $.80. 

19.  If  a  staff  3  ft.  long  cast  a  shadow  5  ft.  in  length,  what 
is  the  height  of  an  object  that  casts  a  shadow  of  46-|  ft.  at 
the  same  time  of  day?  Ans.  28  ft. 

20.  Three  men  hired  a  pasture  for  $63  ;  A  put  in  8  sheep 
7|^  months,  B  put  in  12  sheep  4^  months,  and  0  put  in  15 
sheep  6f  months ;  how  much  must  each  pay  ? 

21.  If  7  bushels  of  wheat  are  worth  10  bushels  of  rye, 
and  5  bushels  of  rye  are  worth  14  bushels  of  oats,  and  6 
bushels  of  oats  are  worth  $3,  how  many  bushels  of  wheat 
will  $30  buy?  A}is.  15. 

22.  If  $480  gain  $84  in  30  months,  what  capital  will  gain 
$21  in  15  months  ?  Ans.  $240. 

23.  How  many  yards  of  carpeting  f  of  a  yard  wide  are 
equal  to  28  yards  |  of  a  yard  wide  ?  Ans.  31  J. 

24.  If  a  footman  travel  130  miles  in  3  days,  when  the  days 
are  14  hours  long,  in  how  many  days  of  7  hours  each  will  he 
travel  390  miles  ?  Ans,  18. 


AIS^ALYSIS.  303 

25.  If  6  men  can  cut  45  cords  of  wood  in  3  days,  how 
many  cords  can  8  men  cut  in  9  days  ?  Ans.  180. 

26.  B's  age  is  1^  times  the  age  of  A,  and  C's  is  2^^  times 
the  age  of  both,  and  the  sum  of  their  ages  is  93  ;  what  is 
the  age  of  each  ?  Ans.  A^s  age,  12  yrs. 

27.  If  A  can  do  as  much  work  in  3  days  as  B  can  do  in 
4|-  days,  and  B  can  do  as  much  in  9  days  as  C  in  12  days, 
and  C  as  much  in  10  days  as  D  in  8,  how  many  days'  work 
done  by  D  are  equal  to  5  days'  done  by  A  ?  Ans.  8. 

28.  The  hour  and  minute  hands  of  a  watch  are  together 
at  12  o'clock,  M. ;  when  will  they  be  exactly  together  the 
third  time  after  this  ? 

OPERATION.  Analysis.     Since  tlie 

12  X  A-  X  3  =  3^^  h.  minute  hand  passes  the 

Ans.  3  h.  16  min.  21-j9^  sec,  p.m.         ^^^  ^^^^^  11  *i"^®s  in 

12  hours,  if  both  are 
together  at  12,  the  minute  hand  will  pass  the  hour  hand  the  first  time 
in  y\  of  12  hours,  or  1-^j  hours  ;  it  will  pass  the  hour  hand  the  second 
time  in  -^^  of  13  hours,  and  the  third  time  in  ^^  of  12  hours,  or  Sf'j 
hours,  which  would  occur  at  16  min.  21^  sec.  past  3  o'clock,  p.  M. 

29.  A  flour  merchant  paid  1164  for  20  barrels  of  flour, 
giving  $9  for  first  quality,  and  $7  for  second  quality;  how 
many  barrels  were  there  of  each  ? 

OPERATION.  Analysis.    If  aU  had  been 

$9  X  20  =  $180  ;  first  quality,  he  would  have 

$180  —  $164  =  $16.  P^i^  ^1^^'  o^  ^1^  ™o^®  *^^  ^« 

Ag $7  =  12  •  ^^  ^^^'     ^^^^  barrel  of  sec- 

Q  , ',  ,  ond  quality  made  a  difference 

Hy^Z=     8  bbl.,  2d  quality.  .^f  ^3  ^  ^^^  ^^^  .  ^^^^  ^^^^^ 

^^  —  8  =  12  bbl.,  ist  quality.  were  as  many  barrels  of  second 

quality  as  $2,  the  difference  in 
the  cost  of  one  barrel,  is  contained  times  in  $16,  etc. 

30.  A  boy  bought  a  certain  number  of  oranges  at  the  rate 
of  3  for  4  cents,  and  as  many  more  at  the  rate  of  5  for  8 
cents  ;  he  sold  them  again  at  the  rate  of  3  for  8  cents,  and 


304  ANALYSIS. 

gained  on  the  whole  108  cents  ;  how  many  oranges  did  he 

buy? 

OPERATION.  Analysis.    For 

1  +  1=  fl;  ^^2z=^,  average  cost        tliose  he   bought 

^-U  =  U  =  H  Cts.,  gain  on  each.  **  ^^^  ^^^  ^^  ^  ^'^^ 

1  A?       TT        JS  4  cents  he  paid  | 

lOo  -T-  1-V  =  yO,  number  of  oranges.  ,          x       i         j 

o             '  of  a  cent  each,  and 

for  those  he  bought  at  the  rate  of  5  for  8  cents  he  paid  f  of  a  cent 

each ;  and  |  +  f  =  ff  cents,  what  he  paid  for  1  of  each  kind,  which 

divided  by  2  gives  f f  cents,  the  average  price  of  all  he  bought.    He 

sold  them  at  the  rate  of  3  for  8  cents,  or  |  cents  each ;  the  diflerence 

between  the  average  cost  and  the  price  he  sold  them  for,  or  |  —  f  f  = 

^1  =  1^  cents,  is  the  gain  on  each  ;  and  he  bought  as  many  oranges  as 

the  gain  on  one  orange  is  contained  times  in  the  whole  gain,  etc. 

31.  A  man  bought  10  bushels  of  wheat  and  25  bushels  of 
com  for  $30,  and  12  bushels  of  wheat  and  5  bushels  of  corn 
for  820  ;  how  much  a  bushel  did  he  give  for  each  ? 

Analysis.  We  may  divide  or 
multiply  either  of  the  expressions 
by  such  a  number  as  will  render 
one  of  the  commodities  purchased, 
alike  in  both  expressions.  In  this 
example  we  divide  the  first  by  5 
to  make  the  numbers  denoting 
the  corn  alike,  (the  same  result 
would  be  produced  by  multiply- 
ing the  second  by  5,)  and  we  have 
the  cost  of  2  bushels  of  wheat  and  5  bushels  of  corn,  equal  to  $6. 
Subtracting  this  from  12  bushels  of  wheat  and  5  bushels  of  com, 
which  cost  $20,  we  find  the  cost  of  10  bushels  of  wheat  to  be  $14 ; 
therefore  the  cost  of  1  bushel  is  yV  ^^  $^4,  or  $1.40.  From  any  one  of 
the  expressions  containing  both  wheat  and  corn,  we  readily  find  the 
cost  of  1  bushel  of  corn  to  be  64  cents. 

32.  A,  B,  and  0  agree  to  build  a  barn  for  $270.  A  and 
B  can  do  the  work  in  16  days,  B  and  C  in  13-J^  days,  and  A 
and  0  in  llf  days.  In  liow  many  days  can  all  do  it  work- 
ing together  ?  In  how  many  days  can  each  do  it  alone  ? 
What  part  of  the  pay  ought  each  to  receive  ? 


urn 
W. 

C. 

Ist  lot,         10 

25     $30 

2d  lot,        12 

5    $20 

^t  ^  5  =  2 

6       $6 

10. 

$14 

1  bu.  W. 

=z  $1.40 

1  bu.  C. 

=  $  .64 

ANALYSIS, 


305 


OPERATION. 

Yg-  =  -go",  what  A  and  B  do  in  1  day. 
•^  =  -^Q,  what  B  and  C  do  in  1  day. 
"eV  ~~  IsV^  what  A  and  C  do  in  1  day. 
A  4-  W  +  A  =  it^  what  A,  B,  and  C  do 

in  2  days. 
^  -T-2  =  -^,  what  A,  B,  and  C  do  in  1  day, 
1  -T-  -^  =  S"!  days,  time  A,  B,  and  C  will  do 
the  whole  work  together. 

A — A  =  "8%  ?  I'^A  —  ^^  da.,  C  alone. 
A~A  =  A?  l-i-^  =  26-|da.,Aalone. 
A — A  =  A  J  1  "^  A  =  ■^^  da.,  B  alone. 
^  X  8-|  =  I",  the  part  of  the  whole  C  did. 
-^  X  8f  =  I",  the  part  of  the  whole  A  did. 
-^  X  Sf  =  f ,  the  part  of  the  whole  B  did. 

$270  Xi  =  1120,  C's  share. 
$270x1=  $90,  A's  share. 
1270  X  f  =  $60,     B'8  share. 


Analysis.  Since 
A  and  B  can  do  the 
work  in  16  days, 
they  can  do  -^^  of  it 
in  1  day ;  B  and  C, 
in  131  or  -^^  days, 
they  can  do  -^  of  it 
in  1  day  ;  A  and  C, 
in  llf  or  ^  days, 
they  can  do  /^  of  it 
in  1  day.  Then  A, 
B,  and  C,  by  work- 
ing 2  days  each,  can 


of  the  work,  and  by 
working  1  day  each 
they  can  do  |  of  |f , 
or  ^^j  of  the  work  ; 
and  it  will  take  them 
as  many  days  work- 
ing together  to  do 
the  whole  work  as  ^^  is  contained  times  in  1,  or  8f  days. 

Now,  if  we  take  what  any  two  of  them  do  in  1  day  from  what  the 
three  do  in  1  day,  the  remainder  will  be  what  the  third  does ;  we  thus 
find  that  A  does  -g%,  B  -g%,  and  C  /^. 

Next,  if  we  denote  the  whole  work  by  1,  and  divide  it  by  the  part 
each  does  in  1  day,  we  have  the  number  of  days  that  it  will  take  each 
to  do  it  alone,  viz. :  A  26|  days,  B  40  days,  and  C  20  days.  And  each 
should  receive  sucb  a  part  of  $270  as  would  be  expressed  by  the  part 
he  does  in  1  day,  multiplied  by  the  number  of  days  he  works,  which 
will  give  to  A  $90,  B  $60,  and  C  $120. 

33.  If  6  oranges  and  7  lemons  cost  33  cents,  and  12 
oranges  and  10  lemons  cost  54  cents,  what  is  the  price  of 
1  of  each?  Ans.  Oranges,  2  cents ;  lemons,  3  cents. 

34.  If  an  army  of  1000  men  have  provisions  for  20  days, 
at  the  rate  of  18  oz.  a  day  to  each  man,  and  they  be  rein- 
forced by  600  men,  upon  what  allowance  per  day  must  each 
man  be  put,  that  the  same  provisions  may  last  30  days  ? 

35.  There  are  54  bushels  of  grain  in  2  bins ;  and  in  one 
bin  are  6  bushels  less  than  J  as  much  as  there  is  in  the 
other  ;  how  many  busliels  in  tlie  larger  bin  ?        Aits.  40. 


306  ANALYSIS. 

36.  The  sum  of  two  numbers  is  20,  and  their  difference 
is  equal  to  ^  of  the  greater  number ;  what  is  the  greater 
number  ?  Ans.  12. 

37.  If  A  can  do  as  much  work  in  2  days  as  C  in  3  days, 
and  B  as  much  in  5  days  as  0  in  4  days  ;  what  time  will  B 
require  to  execute  a  piece  of  work  which  A  can  do  in  6 
weeks  ?  Ans.  11 J  weeks. 

38.  How  many  yards  of  cloth,  |  of  a  yard  wide,  will  line 
36  yards  1 J  yards  wide  ?  Ans.  60. 

39.  How  many  sacks  of  coffee,  each  containing  104  lbs., 
at  lOd.  per  pound  N.  Y.  currency,  will  pay  for  80  yards  of 
broadcloth  at  $3^  per  yard  ?  A7is.  24. 

40.  A  person,  being  asked  the  time  of  day,  replied,  the 
time  past  noon  is  equal  to  -J  of  the  time  to  midnight ;  what 
was  the  hour  ?  Ans.  2  p.m. 

41.  A  market  woman  bought  a  number  of  peaches  at  the 
rate  of  2  for  1  cent,  and  as  many  more  at  the  rate  of  3  for  1 
cent,  and  sold  them  at  the  rate  of  5  for  3  cents,  gaining  55 
cents  ;  how  many  peaches  did  she  buy?  Atis.  300. 

42.  A  can  build  a  boat  in  18  days,  working  10  hours  a  day, 
and  B  can  build  it  in  9  days,  working  8  hours  a  day  ;  in  how 
many  days  can  both  together  build  it,  working  6  hours  a  day  ? 

43.  A  man,  after  spending  ^  of  his  money,  and  J  of  the 
remainder,  had  $10  left ;  how  much  had  he  at  first  ? 

44.  If  30  men  can  perform  a  piece  of  work  in  11  days, 
how  many  men  can  accomplish  another  piece  of  work,  4 
times  as  large,  in  ^  of  the  time  ?  Ans.  600. 

45.  If  16i  lb.  of  coffee  cost  $3  J,  how  much  can  be  bought 
for  $1.25?  Ans.  6^  lb. 

46.  A  man  engaged  to  write  for  20  days,  receiving  $2.50 
for  every  day  he  labored,  and  forfeiting  $1  for  every  day  he 
was  idle ;  at  the  end  of  the  time  he  received  $43 ;  how 
many  days  did  he  labor?  Ans.  18. 

47.  A,  B,  and  C  can  perform  a  piece  of  work  in  12  hours ; 
A  and  B  can  do  it  in  16  hours,  and  A  and  C  in  18  hours  ; 
what  part  of  the  work  can  B  and  C  do  in  9|  hours  ?  Ans.  4. 


ALLIGATION    MEDIAL.  307 

ALLIGATION. 

413,  Alligation  treats  of  mixing  or  compounding  two 
or  more  ingredients  of  different  values.  It  is  of  two  kinds — 
Alligation  Medial  and  Alligation  Alternate, 

414,,  Alligation  Medial  is  the  process  of  finding  the 
average  price  or  quality  of  a  compound  of  several  simple 
ingredients  whose  prices  or  qualities  are  known. 

1.  A  miller  mixes  40  bushels  of  rye  worth  80  cents  a 
bushel,  and  25  bushels  of  corn  worth  70  cents  a  bushel,  with 
15  bushels  of  wheat  worth  $1.50  a  bushel ;  what  is  the  value 
of  a  bushel  of  the  mixture  ? 

Analysis.  Since  40  bushels 
of  rye  at  80  cents  a  bushel  is 
worth  $32,  and  25  bushels  of  corn 
at  70  cents  a  bushel  is  worth 
$17.50,  and  15  bushels  of  wheat 
at  $1.50  a  bushel  is  worth  $22.50, 
^90    An^  therefore  the  entire  mixture,  con- 

sisting of  80  bushels,  is  worth 
$72,  and  one  bushel  is  worth  ^V  of  $72,  or  72  -^  80  =  $.90. 

EuLE.  Divide  the  entire  cost  or  value  of  the  ingredients 
hy  the  sum  of  the  simples. 

Examples  for  Peactice. 

2.  A  wine  merchant  mixes  12  gallons  of  wine,  at  $1  per 
gallon,  with  5  gallons  of  brandy  worth  $1.50  per  gallon,  and 
3  gallons  of  water  of  no  value  ;  what  is  the  worth  of  one 
gallon  of  the  mixture  ?  Ans.  $.975. 

3.  An  innkeeper  mixed  13  gallons  of  water  with  52  gallons 
of  brandy,  which  cost  him  $1.25  per  gallon ;  what  is  the  value 
of  1  gallon  of  the  mixture,  and  what  his  profit  on  the  sale  of 
the  whole  at  6  J  cents  per  gill  ?    Ans.  $1  a  gallon  ;  $65  profit. 

4.  A  grocer  mixed  10  pounds  of  sugar  at  8  cts.  with  12 
pounds  at  9  cts.  and  16  pounds  at  11  cts.,  and  sold  the  mix- 
ture at  10  cents  per  pound ;  did  he  gain  or  lose  by  the  sale, 
and  how  much  ?  Ans.  He  gained  16  cts. 


OPERATION. 

80 

X  40 

=  $32.00 

70 

X  25 

=    17.50 

1.50 

X  15 

=  22.50 

80 

)  72.00 

308  ALLIGATIOJq^    ALTERNATE. 

5.  A  grocer  bought  7J-  dozen  of  eggs  at  12  cents  a  dozen, 
8  dozen  at  10^  cents  a  dozen,  9  dozen  at  11  cents  a  dozen, 
and  lOJ  dozen  at  10  cents  a  dozen.  He  sells  them  so  as  to 
make  50  per  cent,  on  the  cost ;  how  much  did  he  receive 
per  dozen  ?  Ans.  16-J  cents. 

6.  Bought  4  cheeses,  each  weighing  50  pounds,  at  13  cents 
a  pound  ;  10,  weighing  40  pounds  each,  at  10  cents  a  pound ; 
and  24,  weighing  25  pounds  each,  at  7  cents  a  pound ;  I 
sold  the  whole  at  an  average  price  of  9|-  cents  a  pound; 
what  was  my  whole  gain  ?  Ans.  $6. 

41 5.  Alligation  Alternate  is  the  process  of  finding  the 
proportional  quantities  to  be  taken  of  several  ingredients, 
whose  prices  or  qualities  are  known,  to  form  a  mixture  of  a 
required  price  or  quality. 

Case  I. 

416.  To  find  the  proportional  quantity  to  be  used 
of  each  ingredient,  when  the  mean  price  or  quality 
of  the  mixture  is  given. 

1.  What  relative  quantities  of  timothy  seed  worth  $2  a 
bushel,  and  clover  seed  worth  $7  a  bushel,  must  be  used  to 
form  a  mixture  worth  $5  a  bushel  ? 

OPERATION.  Analysis.     Since  on  every  ingre- 

{214     2  )  dient  used  whose  price  or  quality 

ly  \  1     3    j  Ans.  is  less  than  the  mean  rate  there  will 

be  a  gain,  and  on  every  ingredient 
whose  price  or  quality  is  greater  than  the  mean  rate  there  will  be  a 
loss,  and  since  the  gains  and  losses  must  be  exactly  equal,  the  relative 
quantities  used  of  each  should  be  such  as  represent  the  unit  of  value. 
By  selling  one  bushel  of  timothy  seed  worth  |2,  for  $5,  there  is  a  gain 
of  $3  ;  and  to  gain  $1  would  require  ^  of  a  bushel,  which  we  place 
opposite  the  2.  By  selling  one  bushel  of  clover  seed  worth  $7,  for  $5, 
there  is  a  loss  of  $2  ;  and  to  lose  $1  would  require  ^  of  a  bushel,  which 
we  place  opposite  the  7. 

In  every  case,  to  find  the  unit  of  value  we  must  divide  $1  by  the 
gain  or  loss  per  bushel  or  pound,  etc.  Hence,  if,  every  time  we  take 
^  of  a  bushel  of  timothy  seed,  we  take  ^  of  a  bushel  of  clover  seed, 
the  gain  and  loss  will  be  exactly  equal,  and  we  shall  have  I  and  ^  for 
the  proportional  quantities. 


ALLIGATION    ALTERlirATE, 


809 


' 

1 

2 

3 

4 

l^ 

3 

i 

4 

14 

4 

i 

1 

1 

7 

1 

2 

2 

I  10 

i 

3 

3 

If  we  wish  to  express  the  proportional  numbers  in  integers,  we 
may  reduce  these  fractions  to  a  common  denominator,  and  use  their 
numerators,  since  fractions  having  a  common  denominator  are  to  each 
other  as  their  numerators.  ( 378  )  thus,  \  and  \  are  equal  to  f  and  f , 
and  the  proportional  quantities  are  2  bushels  of  timothy  seed  to  3 
bushels  of  clover  seed. 

2.  What  proportions  of  teas  worth  respectively  3,  4,  7  and 
10  shillings  a  pound,  must  be  taken  to  form  a  mixture  worth 
6  shillings  a  pound  ? 

OPERATION.  Analysis.     To  preserve  the 

equality  of  gains  and  losses,  we 
must  always  compare  two  prices 
or  simples,  one  greater  and  one 
less  than  the  mean  rate,  and 
treat  each  pair  or  couplet  as  a 
separate  example.  In  the  given 
example  we  form  two  couplets, 
and  may  compare  either  3  and  10,  4  and  7,  or  3  and  7,  4  and  10. 

We  find  that  \  of  a  lb.  at  3s.  must  be  taken  to  gain  1  shilling,  and 
1^  of  a  lb.  at  10s.  to  lose  1  shilling ;  also,  \  of  a  lb.  at  4s.  to  gain  1  shil- 
ling, and  1  lb.  at  7s.  to  lose  1  shilling.  These  proportional  numbers, 
obtained  by  comparing  the  two  couplets,  are  placed  in  columns  1  and  3. 
If,  now,  we  reduce  the  numbers  in  columns  1  and  2  to  a  common  de- 
nominator, and  use  their  numerators,  we  obtain  the  integral  numbers 
in  columns  3  and  4,  which  being  arranged  in  column  5,  give  the  pro- 
portional quantities  to  be  taken  of  each.* 

It  will  be  seen  that  in  comparing  the  simples  of  any  couplet,  one  of 
which  is  greater,  and  the  other  less  than  the  mean  rate,  the  propor- 
tional number  finally  obtained  for  either  term  is  the  difference  between 
the  mean  rate  and  the  other  term.  Thus,  in  comparing  3  and  10,  the 
proportional  number  of  the  former  is  4,  which  is  the  difference  between 
10  and  the  mean  rate  6  ;  and  the  proportional  number  of  the  latter  is  3, 
which  is  the  difference  between  3  and  the  mean  rate.  The  same  is 
true  of  every  other  couplet.  Hence,  when  the  simples  and  the  mean 
rate  are  integers,  the  intermediate  steps  taken  to  obtain  the  final  pro- 
portional numbers  as  in  columns  1.  2,  3,  and  4,  may  be  omitted,  and 
the  same  results  readily  found  by  taking  the  difference  between  each 
simple  and  the  mean  rate,  and  placing  it  opposite  the  one  with  which 
it  is  compared. 

*  Prof.  A.  B.  Canfield,  of  Oneida  Conference  Seminary,  N.  Y.,  used  this  method  of 
Alligation,  essentially,  in  the  instruction  of  his  classes  as  early  as  1846,  and  he  was 
doubtless  tho  author  of  it. 


310  ALLIGATION    ALTERNATE. 

Rule.  I.  Write  the  several  prices  or  qualities  in  a  column, 
and  the  mean  price  or  quality  of  the  mixture  at  the  left. 

II.  For7)i  couplets  by  comparing  any  price  or  quality  less, 
with  one  that  is  greater  than  the  mean  rate,  placing  the  pari 
which  must  be  used  to  gain  1  of  the  mean  rate  opposite  the  less 
simple,  and  the  part  that  must  be  used  to  lose  1  opposite  the 
greater  simple,  and  do  the  same  for  each  simple  in  every 
couplet, 

III.  If  the  proportional  numbers  are  fractional,  they  may 
be  reduced  to  integers,  and  if  tivo  or  more  stand  in  the  same 
horizontal  line,  they  must  be  added;  the  final  results  will  be 
the  proportional  quantities  required, 

1.  If  the  numbers  in  any  couplet  or  column  have  a  common  factor,  it  may  be 
rejected. 

2.  We  may  also  multiply  the  numbers  in  any  couplet  or  column  by  any  multiplier 
we  choose,  without  affecting  the  equality  of  the  gains  and  losses,  and  thus  obtain 
an  indefinite  number  of  results,  any  one  of  which  being  taken  will  give  a  correct 
final  result. 

Examples  for  Practice. 

3.  A  grocer  has  sugars  worth  10  cents,  11  cents,  and  14 
cents  per  pound ;  in  what  proportions  may  he  mix  them  to 
form  a  mixture  worth  12  cents  per  pound  ? 

Ans.  1  lb.  at  10  cts.,  and  2  lbs.  at  11  and  14  cts. 

4.  What  proportions  of  water  at  no  value,  and  wine  worth 
$1.20  a  gallon,  must  be  used  to  form  a  mixture  worth  90  cents 
a  gallon  ?  Ans,  1  gal.  of  water  to  3  gals,  of  wine. 

5.  A  farmer  had  sheep  worth  $2,  $2|^,  $3,  and  $4  per 
head ;  what  number  could  he  sell  of  each,  and  realize  an 
average  price  of  $2f  per  head  ? 

Ajis.  5  of  the  1st  kind,  and  1  each  of  the  2d  and  3d,  ana 
3  of  the  4th  kind. 

6.  What  relative  quantities  of  alcohol  80,  84,  87,  94,  and 
96  per  cent,  strong  must  be  used  to  form  a  mixture  90  per 
cent,  strong? 

Ans.  6  of  the  first  two  kinds,  4  of  the  3d,  3  of  the  4th, 
and  16  of  the  5  th. 


alligatioit   alternate.  811 

Case  IL 
417.  When  the  quantity  of  one  of  the  simples  is 
limited. 

1.  A  miller  has  oats  worth  30  cents,  com  worth  45  cents, 
and  barley  worth  84  cents  per  bushel ;  he  desires  to  form  a 
mixture  worth  60  cents  per  bushel,  and  which  shall  contain 
40  bushels  of  com ;  how  many  bushels  of  oats  and  barley 
must  he  take  ? 

OPEEATION.  Analysis.    By 

the  same  process 

60  H  45  A         8      8   40   h  Ans.      ^^  ^^  ^'"^^  ^  ^® 


po 

J  45 
184 


A 

4 

4 

h 

8 

8 

A 

^ 

5 

5 

10 

find    the    proper- 
84    A    ^     5    5      10    OO  J  ^.^^^1    quantities 

of  each  to  be  4  bushels  of  oats,  8  of  com,  and  10  of  barley.  But  we 
wish  to  use  40  bushels  of  corn,  which  is  5  times  the  proportional 
number  8 ;  and  to  preserve  the  equality  of  gain  and  loss  we  must  take 
5  times  the  proportional  quantity  of  each  of  the  other  simples,  or 
5  X  4  =  20  bushels  of  oats,  and  5  x  10  =  50  bushels  of  barley. 

Rule.  Find  the  proportional  quantifies  as  in  Case  L 
Divide  the  given  quantity  by  the  proportional  quantity  of  the 
same  ingredient,  and  multiply  each  of  the  other  proportional 
quantities  by  the  quotient  thus  obtained. 

Examples  foe  Pkactice. 

2.  A  merchant  has  teas  worth  40,  60,  75,  and  90  cents  per 
pound ;  how  many  pounds  of  each  must  he  use  with  20 
pounds  of  that  worth  75  cents,  to  form  a  mixture  at  80  cents  ? 

Ans.  20  lbs.  each  of  the  first  three  kinds,  and  130  lbs.  of 
the  fourth. 

3.  A  farmer  bought  24  sheep  at  $2  a  head;  how  many 
must  he  buy  at  $3  and  $5  a  head,  that  he  may  sell  the  whole 
at  an  average  price  of  $4  a  head,  without  loss  ? 

Ans.  24  at  $3,  and  72  at  $5. 

4.  How  much  alcohol  worth  60  cents  a  gallon,  and  how 
much  water,  must  be  mixed  with  180  gallons  of  rum  worth 
$1.30  a  gallon,  that  the  mixture  may  be  worth  90  cents  a 
gallon  ?  Ans.  60  gallons  each  of  alcohol  and  water. 


813 


ALLIGATION    ALTERNATE 


5.  How  many  acres  of  land  worth  35  dollars  an  acre  must 
be  added  to  a  farm  of  75  acres,  worth  $50  an  acre,  that  the 
average  value  may  be  $40  an  acre  ?  Ans.  150  acres. 

6.  A  merchant  mixed  80  pounds  of  sugar  worth  6  J  cents 
per  pound  with  some  worth  8^  cents  and  10  cents  per  pound, 
so  that  the  mixture  was  worth  7^  cents  per  pound;  how 
much  of  each  kind  did  he  use  ? 

Case  III. 
418.  "When  the  quantity  of  the  whole  componnd 
is  limited. 

1.  A  grocer  has  sugars  worth  6  cents,  7  cents,  12  cents, 
and  13  cents  per  pound.  He  wishes  to  make  a  mixture  of 
120  pounds  worth  10  cents  a  pound ;  how  many  pounds  of 
each  kind  must  he  use  ? 

OPERATION.  Analysis.    By  Case  1 

we  find  the  proportional 
quantities  of  each  to  be 
10  ^^I  y  I~«"  8  lbs.  at  6  cts.,  2  lbs.  at 

7  cts.,  3  lbs.  at  12  cts., 
and  4  lbs.  at  13  cts.  By 
adding  the  proportional 
quantities,  we  find  that 
the  mixture  would  be  but  12  lbs.  while  the  required  mixture  is  120,  or 
10  times  13.  If  the  whole  mixture  is  to  be  10  times  as  much  as  the 
sum  of  the  proportional  quantities,  then  the  quantity  of  each  simple 
used  must  be  10  times  as  much  as  its  respective  proportional,  which 
would  require  30  lbs.  at  6  cts.,  20  lbs.  at  7  cts.,  30  lbs.  at  12  cts.,  and 
40  lbs.  at  13  cts. 

Rule.  Find  the  proportional  numbers  as  in  Case  I.  Di- 
vide the  given  quantity  by  the  sum  of  the  proportional  quan- 
tities, and  multiply  each  of  the  proportional  quantities  by  the 
quotient  thus  obtained. 

Examples  for  Practice. 

2.  A  farmer  sold  170  sheep  at  an  average  price  of  14: 
shillings  a  head  ;  for  some  he  received  9s.,  for  some  12s.,  for 
some  18s.,  and  for  others  20s. ;  how  many  of  each  did  he 
BcU?    Ans,  60  at  9s.,  40  at  12s.,  20  at  18s.,  and  50  at  20s. 


'    6 

J 

3 

3 

30 

.      ^ 

i 

2 

2 

20 

12 

i 

3 

3 

30 

.13 

i 

4 

4 

40 

12 

120 

IKVOLUTION.  313 

3.  A  jeweler  melted  together  gold  16,  18,  21,  and  24 
carats  fine,  so  as  to  make  a  compound  of  51  ounces  22  carats 
fine  ;  how  much  of  each  sort  did  he  take  ?  Ans,  6  ounces 
each  of  the  first  three,  and  33  ounces  of  the  last. 

4.  A  man  bought  210  bushels  of  oats,  corn,  and  wheat, 
and  paid  for  the  whole  $178.50  ;  for  the  oats  he  paid  I  J,  for 
the  corn  If,  and  for  the  wheat  $1 J  per  bushel ;  how  many- 
bushels  of  each  kind  did  he  buy?  Ans*  78  bushels  each  of 
oats  and  corn,  and  54  bushels  of  wheat. 

5.  A,  B,  and  C  are  under  a  joint  contract  to  furnish  6000 
bushels  of  corn,  at  48  cts.  a  bushel ;  A's  com  is  worth  45  cts., 
B's  51  cts.,  and  C's  54  cts. ;  how  many  bushels  must  each 
put  into  the  mixture  that  the  contract  may  be  fulfilled  ? 

6.  One  man  and  3  boys  received  $84  for  56  days'  labor ;  the 
man  received  $3  per  day,  and  the  boys  $J,  $|,  and  |1|  re- 
spectively ;  how  many  days  did  each  labor  ?  Ans.  The  man 
16  days,  and  the  boys  24,  4,  and  12  days  respectively. 

INYOLUTIOlSr. 

419.  A  Power  is  the  product  arising  from  multiplying 
a  number  by  itself,  or  repeating  it  several  times  as  a  factor ; 
thus,  in  2  X  2  X  2  =  8,  the  product,  8,  is  a  power  of  2. 

420.  The  Exponent  of  a  power  is  the  number  dent) ting 
how  many  times  the  factor  is  repeated  to  produce  the  power, 
and  is  written  above  and  a  little  to  the  right  of  the  factor ; 
thus,  2  X  2  X  2  is  written  28,  in  which  3  is  the  exponent 
Exponents  likewise  give  names  to  the  powers,  as  will  be 
seen  in  the  following  illustrations  : 

3  =  31  =    3,  the  first  power  of  3  ; 

3x3  =  33  =    9,  the  second  power  of  3  ; 

3x3x3     =  33  =  27,  the  third  power  of  3. 

431.  The  Square  of  a  number  is  its  second  power. 

422.  The  Cube  of  a  number  is  its  third  power. 

423.  Involution  is  the  process  of  raising  a  number  to 
a  given  power. 

R.P.  14 


314  EVOLUTION. 

424.  A  Perfect  Power  is  a  number  that  can  be  exactly 
produced  by  the  involution  of  some  number  as  a  root ;  thus, 
25  and  32  are  perfect  powers,  since  25  =  5  x  5,  and  32  =  2 
X2x2x2x2. 
1.  What  is  the  cube  of  15  ? 

OPERATION.  Analysis.    We  multiply  15  by 

'6  X  15  X  15  =  3375,  Ans»       15,  and  the  product  by  15,  and  ob- 
tain 3375,  which  is  the  3d  power, 
or  cube  of  15,  since  15  has  been  taken  3  times  as  a  factor. 

EuLE.  Multiply  the  number  hy  itself  as  many  times,  less  1, 
as  there  are  units  in  the  exponent  of  the  required  power. 

Examples  for  Practice. 

2.  What  is  the  square  of  25  ?  Ans.  625. 

3.  What  is  the  square  of  135?  Ans.  18225. 

4.  What  is  the  cube  of  72  ?  Ans.  373248. 

5.  What  is  the  4th  power  of  24?  Ans.  331776. 

6.  Eaise  7.2  to  the  third  power.  Ans.  373.248. 

7.  InYolve  1.06  to  the  4th  power.  Ans.  1.26247696. 

8.  Involve  .12  to  the  5th  power.  Ans.  .0000248832. 

9.  Involve  1.0002  to  the  2d  power.  Ans.  1.00040004. 
10.  What  is  the  cube  of  |  ? 

OPERATION. 

2       2       2       2x2x2       23         8 
x-=-x 


5       5  '    5      6x5x5      5»       125 
It  is  evident  from  the  above  operation,  that 
A  common  fraction  may  he  raised  to  any  'power,  ly  raising 
each  of  its  terms,  separately,  to  the  required  power. 

11.  What  is  the  square  of  f  ?  Ans.  ■^. 

12.  What  is  the  cube  of  \l  ?  Ans.  f^H. 

13.  Raise  24^  to  the  2d  power.  Ans.  612^. 

EYOLUTIOlSr. 

425.  A  Boot  is  a  factor  repeated  to  produce  a  power , 
thus,  in  the  expression  5x5x5=:  125,  5  is  the  root  from 
which  the  power,  125,  is  produced. 


SQUARE    ROOT.  315 

426.  Evolution  is  the  process  of  extracting  the  root 
of  a  number  considered  as  a  power,  and  is  the  reyerse  of 
Involution. 

427.  The  Radical  Sign  is  the  character,  y',  which, 
placed  before  a  number,  denotes  that  its  root  is  to  be  extracted. 

428.  The  Index  of  the  root  is  the  figure  placed  above 
the  radical  sign,  to  denote  what  root  is  to  be  taken.  When 
no  index  is  written,  the  index  2  is  always  understood. 

429.  A  Surd  is  the  indicated  root  of  an  imperfect  power. 
4.30.  Roots  are  named  from  the  corresponding  powers, 

as  will  be  seen  in  the  following  illustrations  : 
The  square  root  of  9  is  3,  written  a/9~=:  3. 
The  cube  root  of  27  is  3,  written  \^W=  3. 
The  fourth  root  of  81  is  3,  written  v^sTrzi  3. 

431.  Any  number  whatever  may  be  considered  a  power 
whose  root  is  to  be  extracted ;  but  only  the  perfect  powers 
can  have  exact  roots. 

SQUAEE    ROOT. 

432.  The  Square  Root  of  a  number  is  one  of  the  two 

equal  factors  that  produce  the  number;  thus  the  square 
root  of  49  is  7,  for  7  x  7  =  49. 

433.  In  extracting  the  square  root,  the  first  thing  to  be 
determined  is  the  relative  number  of  places  in  a  given  num- 
ber and  its  square  root.  The  law  governing  this  relation  is 
exhibited  in  the  following  examples  : 


Roots. 

Squares. 

Roots. 

Squares. 

1 

1 

1 

1 

9 

81 

10 

1,00 

99 

98,01 

100 

1,00,00 

999 

99,80,01 

1000 

1,00,00,00 

From  these  examples  we  perceive 

1st.  That  a  root  consisting  of  1  place  may  have  1  or  2 
places  in  the  square. 

2d.  That  in  all  cases  the  addition  of  1  place  to  the  root 
adds  2  places  to  the  square.    Hence, 


316  EVOLUTION. 

If  we  point  off  a  number  into  two-figure  periods,  commen- 
cing at  the  right  hand,  the  number  of  full  periods  and  the 
left  hand  full  or  partial  period  will  indicate  the  number  of 
places  in  the  square  root ;  the  highest  period  answering  to 
the  highest  figure  of  the  root, 

434.  1.  What  is  the  length  of  one  side  of  a  square  plat 
containing  an  area  of  5417  sq.  ft. 

OPEEATION.  Analysis.     Since  the  given  figure  is 

54,17  I  73.6  a  square,  its  side  will  be  tlie  square  root 

A(^  "  of  its  area,  which  we  will  proceed  to  com- 

pute.   Pointing  off  the  given  number,  the 


140      517  2  periods  show  that  there  will  be  two  in- 

143      429  tegral  figures,  tens  and  units,  in  the  root. 

146  0     88  00  "^^^  *^^^  ^^  *^®  ^^*  must  be  extracted 
-t  Ar>  n     o»v/^/>  from  tlio  first  or  left  hand  period,  54  hun- 
146.6     87.96  j    j       mi            ^    .              •     fr^  i 
dreds.     The  greatest  square  in  54  hun- 

^  dreds  is  49  hundreds,  the  square  of  7  tens ; 

we  therefore  write  7  tens  in  the  root,  at 
the  right  of  the  given  number. 
Since  the  entire  root  is  to  be  the  side  of  a  square,  let  us  form  a 
square  (Fig.  I),  the  side  of  which  is  70  feet  long. 
The  area  of  this  square  is  70  x  70  =  4900  sq.  ft, 
which  we  subtract  from  the  given  number.  This 
is  done  in  the  operation  by  subtracting  the 
square  number,  49,  from  the  first  period,  54, 
and  to  the  remainder  bringing  down  the  second 
period,  making  the  entire  remainder  517. 
If  we  now  enlarge  our  square  (Fig.  I)  by  the  addition  of  517  square 
feet,  in  such  a  manner  as  to  preserve  the  square  form,  its  size  will  be 
that  of  the  required  square.  To  preserve  the  square  form,  the  addition 
must  be  so  made  as  to  extend  the  square  equally  in  two  directions ; 
it  will  therefore  be  composed  of  2  oblong  figures  at  the  sides,  and  a 
little  square  at  the  corner  (Fig.  II).  Now,  the  width  of  this  addition 
will  be  the  additional  length  to  the  side  of  the  square,  and  conse- 
quently the  next  figure  in  the  root.  To  find  width  we  divide  square 
contents,  or  area,  by  length.  But  the  length  of  one  side  of  the  little 
square  cannot  be  found  till  the  width  of  the  addition  be  determined, 
because  it  is  equal  to  this  width.  We  will  therefore  add  the  lengths 
of  the  2  oblong  figures,  and  the  sum  will  bo  sufficiently  near  th© 
whole  length  to  be  used  as  a  trial  divisor. 


SQUAKE    ROOT. 


317 


Fig.  IT. 


ec 

3 

70 

70 

70 

3 

j         Trial  Divisor  =  140. 

1 

i 

1          Complete  Divisor  =  143. 

Each  of  the  oblong  figures  is  equal  in  length  to  the  side  of  the 
square  first  formed  ;  and  their  united  length 
is  70  +  70  =  140  ft.  (Fig.  III).  This  num- 
ber is  obtained  in  the  operation  by  doubling 
the  7  and  annexing  1  cipher,  the  result  being 
written  at  the  left  of  the  dividend.  Dividing 
517,  the  area,  by  140,  the  approximate  length, 
we  obtain  3,  the  probable  width  of  the  addi- 
tion, and  second  figure  of  the  root.  Since  3  is 
also  the  side  of  the  little  square,  we  can  now 
find  the  entire  length  of  the  addition,  or  the  complete  divisor,  which 

is  70  4-  70  +  3  =  143  (Fig.  III). 
'^'      '  This  number  is  found  in  the  oper- 

ation by  adding  3  to  the  trial 
divisor,  and  writing  the  result  un- 
derneath. Multiplying  the  com- 
plete divisor,  143,  by  the  trial 
quotient  figure,  3,  and  subtracting 
the  product  from  the  dividend,  wo 
obtain  another  remainder  of  88  square  feet.  With  this  remainder, 
for  the  same  reason  as  before,  we  must  proceed  to  make  a  new 
enlargement ;  and  we  bring  down  two  decimal  ciphers,  because  the 
next  figure  of  the  root,  being  tenths,  its  square  will  be  hundredths. 
The  trial  divisor  to  obtain  the  width  of  this  new  enlargement,  or  the 
next  figure  in  the  root,  will  be,  for  the  same  reason  as  before,  twice 
73,  the  root  already  found,  with  one  cipher  annexed.  But  since  the 
7  has  already  been  doubled  in  the  operation,  we  have  only  to  double 
the  last  figure  of  the  complete  divisor,  143,  and  annex  a  cipher,  to 
obtain  the  new  trial  divisor,  146.0.  Dividing,  we  obtain  .6  for  the 
trial  figure  of  the  root ;  then  proceeding  as  before,  we  obtain  146.6 
for  a  complete  divisor,  87.96  for  a  product ;  and  there  is  still  a  re- 
mamdei  of  .04.  Hence,  the  side  of  the  given  square  plat  is  73.6  feet, 
nearly     From  this  example  and  analysis  we  deduce  the  following 

Rule.  I.  Point  off  the  given  numher  into  periods  of  two 
figures  eacli^  counting  from  unifs  place  toward  the  left  and 
right, 

11.  Find  the  greatest  square  number  in  the  left  hand  period, 
and  ivrite  its  root  for  the  first  figure  in  the  root ;  subtract  the^ 
square  number  from  the  left  hand  period,  and  to  tlie  remain- 
der bring  down  the  next  period  for  a  dividend. 


318  EVOLUTION". 

III.  At  the  left  of  the  dividend  write  twice  the  first  figure 
of  the  root,  and  annex  o?ie  cipher,  for  a  trial  divisor  ;  divide 
the  dividend  by  the  trial  divisor,  and  write  the  quotient  for  a 
trial  figure  in  the  root. 

IV.  Add  the  trial  figure  of  the  root  to  the  trial  divisor  for 
a  complete  divisor;  multiply  the  complete  divisor  hy  the  trial 
figure  in  the  root,  and  suUract  the  product  from  the  divi- 
dend, and  to  the  remainder  bring  down  the  next  'period  for  a 
new  dividend, 

V.  To  the  last  complete  divisor  add  the  last  figure  of  the 
root,  and  to  the  sum  annex  one  cipher,  for  a  netv  trial  divisor, 
with  which  proceed  as  before. 

1.  If  at  any  time  the  product  be  greater  than  the  dividend,  diminish  the  trial  figure 
of  the  root,  and  correct  the  erroneous  work. 

2.  If  a  cipher  occur  in  the  root,  annex  another  cipher  to  the  trial  divisor,  and 
another  period  to  the  dividend,  and  proceed  as  before. 

Examples  for  Practice. 

2.  What  is  the  square  root  of  406457.2516  ? 

OPERATION. 

40,64,57.25,16  |  637.54,  Ans. 
36 

Trill  divisor,    120 
Complete 

Trial 
Complete 

Trial 
Complete 

Trial 
Complete 

8.  The  decimal  points  in  the  work  may  be  omitted,  care  being  taken  to  point  ofT 
in  the  root  according  to  the  number  of  decimal  periods  used. 

4.  The  pupil  will  acquire  greater  facility,  and  secure  gi-eater  accuracy,  by  keeping 
units  of  like  order  under  each  other,  and  each  divisor  opposite  the  corresponding 
dividend,  by  the  use  of  the  lines,  as  shown  in  the  operation. 

3.  What  is  the  square  root  of  676  ?  Ans.  24. 


120 
123 

464 
369 

1260 
1267 

9557 
8869 

1274.0 
1274.5 

688.25 
637.25 

1275.00 
1275.04 

51.0016 
51.0016 

SQUARE     ROOT.  319 

4.  What  is  the  square  root  of  6561  ?  Ans.  81. 

5.  What  is  the  square  root  of  444889  ?  Ans,  667. 

6.  What  is  the  square  root  of  994009  ?  A^is.  997. 

7.  What  is  the  square  root  of  29855296  ?  Ans.  5464. 

8.  What  is  the  square  root  of  3486784401  ?  Ans.  59049. 

9.  What  is  the  square  root  of  54819198225  ? 

5.  The  cipher  in  the  trial  divisor  may  be  omitted,  and  its  place,  after  division,  oo 
cupied  by  the  trial  root  figure,  thus  forming  in  succession  only  complete  divisors. 

10.  What  is  the  square  root  of  2  ? 

2.        I  1.4:U2  +  ,Ans. 
1 


24 

100 
96 

281 

400 
281 

2824 

11900 
11296 

28282 

60400 
56564 

11.  Extract  the  square  roots  of  the  following  numbers 


V3  =  1.7320508 + 
\/5  =  2.2360679 + 
a/6  =  2.4494897 + 


V^    =2.6457513  + 

'v/8_  =  2.8284271  + 
a/10  =  3.1622776  + 


12.  What  is  the  square  root  of  .00008836  ?      Ans.  .0094. 

13.  What  is  the  square  root  of  .0043046721  ?  Ans.  .06561. 

6.  The  square  root  of  a  common  fraction  may  be  obtained  by  extracting  the  square 
roots  of  the  numerator  and  denominator  separately,  provided  the  terms  are  perfect 
squares ;  otherwise,  the  fraction  may  first  be  reduced  to  a  decimal. 

7.  Mixed  numbers  maybe  reduced  to  the  decimal  form  before  extracting  the  root; 
or,  if  the  denominator  of  the  fraction  is  a  perfect  square,  to  an  improper  fraction. 

14.  Extract  the  square  root  of  -gS^.  Ans.  ff. 

15.  Extract  the  square  root  of  ^^.  Ans.  -J. 

16.  Extract  the  square  root  of  |.  Ans.  .816i96 -\- . 

17.  Extract  the  square  root  of  17|.  ^ws.  4.1683 -t-. 


820 


EVOLUTIOIf 


Applicatioits. 

435.  An  Angle  is  the  opening  between  two  lines  that 
meet  each  other  ;  thus,  the  two  lines,  A  B  and  A  C  meet- 
ing, form  an  angle  at  A. 

436.  A  Triangle  is  a  figure  having  three 
sides  and  three  angles,  as  A,  B,  0. 

43*7.  A  Right-Angled  Triangle  is  a  tri- 
angle having  one  right  angle,  as  at  C. 

438.  The  Base  is  the  side  on  which  it 
stands,  as  A,  0. 

439.  The  Perpendicular  is   the   side 
forming  a  right  angle  with  the  base,  as  B,  0. 

440.  The  Hypotenuse  is  the  side  opposite  the  right 
angle,  as  A,  B. 

441.  Those  examples  given  below,  which  relate  to  trian- 
gles and  circles,  may  be  solved  by  the  use  of  the  two  fol- 
lowing principles,  which  are  demonstrated  in  geometry. 

1st.  The  square  of  the  hypotenuse  of  a  right-angled  trian- 
gle is  equal  to  the  sum  of  the  squares  of  the  other  tivo  sides. 

2d.  T/ie  areas  of  two  circles  are  to  each  other  as  the  squares 
of  their  radii,  diameters,  or  circwnferences, 

1.  The  two  sides  of  a  right-angled  triangle  are  3  and  4 
feet ;  what  is  the  length  of  the  hypotenuse  ? 

Analysis.    Squaring 
OPERATION.  the  two  sides  and  add- 

38  =    9,  square  of  one  side.  ing,  we  find  the  sum  to 

42  =  16,  square  of  the  other  side.      ^^  ^5 ;  and  since  the  sum 

■^s  J.  1  i  is  equal  to  the  square  of 

25,  square  of  hypotenuse.  .^    \      ^ 
'    ^                   ''^  the  hypotenuse,  we  ex- 

V25  =  5,  Ans.  tract    the  square  root, 

and  obtain  5  feet,  the 

hypotenuse.    Hence, 

To  find  the  hypotenuse.  Add  the  squares  of  the  two  sides, 
and  extract  the  square  root  of  the  sum. 

To  find  either  of  the  shorter  sides.  Suhtract  the  square 
of  the  given  side  from  the  square  of  the  hypotenuse,  and  ex- 
tract the  square  root  of  the  remainder. 


square  root.  321 

Examples  for  Practice. 

2.  If  an  army  of  55225  men  be  drawn  up  in  the  form  of  a 
square,  how  many  men  will  there  be  on  a  side  ?   Ans.  235. 

3.  A  man  has  200  yards  of  carpeting  1^  yards  wide  ;  what 
is  the  length  of  one  side  of  the  square  room  which  this  car- 
pet will  cover  ?  Ans.  45  feet. 

4.  How  many  rods  of  fence  will  be  required  to  inclose  10 
acres  of  land  in  the  form  of  a  square  ?         Ans,  160  rods. 

5.  The  top  of  a  castle  is  45  yards  high,  and  the  castle  is 
surrounded  by  a  ditch  60  yards  wide  ;  required  the  length 
of  a  rope  that  will  reach  from  the  outside  of  the  ditch  to 
the  top  of  the  castle.  Ans.  75  yards. 

6.  Eequired  the  height  of  a  May-pole,  which  being  broken 
39  feet  from  the  top,  the  end  struck  the  ground  15  feet  from 
the  foot.  Ans.  75  feet.  . 

7.  A  ladder  40  feet  long  is  so  placed  in  a  street,  that 
without  being  moved  at  the  foot,  it  will  reach  a  window  on 
one  side  33  feet,  and  on  the  other  side  21  feet,  from  the 
ground ;  what  is  the  breadth  of  the  street  ?   Ans,  56. 64+ ft. 

8.  A  ladder  52  feet  long  stands  close  against  the  side  of  a 
building  ;  how  many  feet  must  it  be  drawn  out  at  the  bot- 
tom, that  the  top  may  be  lowered  4  feet?     Ans,  20  feet. 

9.  Two  men  start  from  one  comer  of  a  park  one  mile 
square,  and  travel  at  the  same  rate.  A  goes  by  the  walk 
around  the  park,  and  B  takes  the  diagonal  path  to  the  opposite 
corner,  and  turns  to  meet  A  at  the  side.  How  many  rods 
from  the  comer  will  the  meeting  take  place?  Ans.  93. 7  -|-  rods. 

10.  A  room  is  20  feet  long,  16  feet  wide,  and  12  feet  high; 
what  is  the  distance  from  one  of  the  lower  comers  to  the 
opposite  upper  comer  ?  Ans.  28.284271+  feet. 

11.  It  requires  63.39  rods  of  fence  to  inclose  a  circular 
lield  of  2  acres ;  what  length  will  be  required  to  inclose  3 
acres  in  circular  form  ?  Ans.  77.63+  rods. 

12.  The  radius  of  a  certain  circle  is  5  feet ;  what  will  be 
the  radius  of  another  circle  containing  twice  the  area  of  the 
first  ?  ,^  Ans.  7.07106+  feet. 


322  EVOLUTIOK. 

CUBE    ROOT. 

442.  The  Cube  Koot  of  a  number  is  one  of  the  three 
equal  factors  that  produce  the  number.  Thus,  the  cube 
root  of  27  is  3,  since  3  x  3  x  3  =  27. 

443.  In  extracting  the  cube  root,  the  first  thing  to  be 
determined  is  the  relative  number  of  places  in  a  cube  and 
its  root.  The  law  governing  this  relation  is  exhibited  in 
the  following  examples  : 


Roots. 

Cubes. 

Roots. 

Cnbes. 

1 

1 

1 

1 

9 

729 

10 

1,000 

99 

907,299 

100 

1,000,000 

999 

997,002,999 

1000 

1,000,000,000 

From  these  examples,  we  perceive, 

1st.  That  a  root  consisting  of  1  place  may  have  from  1  to 
3  places  in  the  cube. 

2d.  That  in  all  cases  the  addition  of  1  place  to  the  root 
adds  three  places  to  the  cube. 

If  we  point  off  a  number  into  three-figure  periods,  com- 
mencing at  the  right  hand,  the  number  of  full  periods  and 
the  left  hand  full  or  partial  period  will  indicate  the  number 
of  places  in  the  cube  root,  the  highest  period  answering  to  the 
highest  figure  of  the  root, 

444.  1.  What  is  the  length  of  one  side  of  a  cubical 
block  coctaining  413494  solid  inches  ? 

OPERATION— COMMENCED  ANALYSIS.    Since  the  block  is  a 

413494  I  74  cube,  its  side  will  be  the  cube  root 

g^g  of  its  solid  contents,  which  we  will 

proceed  to  compute.     Pointing  off 

14700     70494  ^^iq  given  number,  the  two  periods 

show  that  there  will  be  two  figure^, 
tens  and  units,  in  the  root.  The  tens  of  the  root  must  be  extracted 
from  the  first  period,  413  thousands.  The  greatest  cube  in  413  thou- 
sands is  343  thousands,  the  cube  of  7  tens  ;  we  therefore  write  7  tens 
in  the  root  at  the  right  of  the  given  number. 


CUBE    ROOT. 


323 


Since  the  entire  root  is  to  be  the  side  of  a  cube,  let  us  form  a 

cubical  block  (Fig.  I),  the  side 
of  which  is  70  inches  in  length. 
The  contents  of  this  cube  are 
70  X  70  X  70  =  343,000  solid 
inches,  which  we  subtract  from 
the  given  number.  This  is  done 
in  the  operation  by  subtracting 
the  cube  number,  343,  from  the 
first  period,  413,  and  to  the  re- 
mainder bringing  down  the  sec- 
ond period,  making  the  entire 
remainder  70494. 

If  we  now  enlarge  our  cubical 
block,  (Fig.  I),  by  the  addition  of  70494  solid  inches,  in  such  a  man- 
ner as  to  preserve  the  cubical  form,  its  size  will  be  that  of  the  required 
block.  To  preserve  the  cubical  form,  the  addition  must  be  made  upon 
three  adjacent  sides  or  faces.  The  addition  will  therefore  be  com- 
posed of  3  flat  blocks  to  cover  the  3  faces,  (Fig.  II) ;  3  oblong  blocks 
to  fill  the  vacancies  at  the  edges,  (Fig.  Ill) ;  and  1  small  cubical  block 
to  fill 'the  vacancy  at  the  corner,  (Fig.  IV.)  Now,  the  thickness  of  this 
enlargement  will  be  the  additional  length  of  the  side  of  the  cube, 
and,  consequently,  the  second  figure  in  the  root.  To  find  thickness, 
we  may  divide  solid  contents  by  surface,  or  area.     But  the  area  of 

the  3  oblong  blocks  and  lit- 
tle cube  cannot  be  found 
till  the  thickness  of  the  ad- 
dition be  determined,  be- 
cause their  common  breadth 
is  equal  to  this  thickness. 
We  will  therefore  find  the 
area  of  the  three  flat  blocks, 
which  is  sufficiently  near 
the  whole  area  to  be  used 
as  a  trial  divisor.  As  these 
are  each  equal  in  length  and 
breadth  to  the  side  of  the 
cube  whose  faces  they  cover, 
the  whole  area  of  the  three 
is  70  X  70  X  3  =  14700  square  inches.  This  number  is  obtained  in  the 
operation  by  annexing  2  ciphers  to  three  times  the  square  of  7 ;  the  re- 
sult being  written  at  the  left  hand  of  the  dividend.    Dividing,  we  obtain 


324 


EVOLUTION 


OPERATION — CONTINUED. 

413494  I  74 
I.     II.  343        


4,  the  probable  thickness  of  the  addition,  and  second  figure  of  the 

root.  With  this  assumed  figure, 
we  will  complete  our  divisor  by 
adding  the  area  of  the  4  blocks, 
before  undetermined.  The  3  ob- 
long blocks  are  each  70  inches 
long ;  and  the  little  cube,  being 
equal  in  each  of  its  dimensions 
to  the  thickness  of  the  addition, 
must  be  4  inches  long.  Hence, 
their  united  length  is  70  +  70 
+  70  +  4  =  214.  This  number 
is  obtained  in  the  operation  by 
multiplying  the  7  by  3,  and  an- 
nexing the  4  to  the  product,  the 
result  being  written  in  column 
I,  on  the  next  line  below  the 
trial  divisor.  Multiplying  314, 
the  length,  by  4,  the  common 
width,  we  obtain  856,  the  area  of 
the  four  blocks,  which  added  to 
14700,  the  trial  divisor,  makes 
15556,  the  complete  divisor ;  and 
multiplying  this  by  4,  the  second 
figure  in  the  root,  and  subtract- 
ing the  product  from  the  divi- 
dend, we  obtain  a  remainder  of 
8270  solid  inches.  With  this  re- 
mainder, for  the  same  reason  as 
before,  we  must  proceed  to  make 
a  new  enlargement.  But  since 
we  have  already  two  figures  in 
the  root,  answering  to  the  two 
periods  of  the  given  number, 
the  next  figure  of  the  root  must 
be  a  decimal ;  and  we  therefore 
annex  to  the  remainder  a  period 
of  three  decimal  ciphers,  mak- 
ing 8270.000  for  a  new  dividend. 
The  trial  divisor  to  obtain  the 
thickness  of  this  second  enlarge- 
ment, or  the  next  figure  of  the  root,  will  be  the  area  of  three  new  flat 
l)!ocks  to  cover  the  three  sides  of  the  cube  already  formed  ;  and  tb.i." 


214  856 


14700 
15556 


70494 
62224 


8270.000 


J 


<:;UBB     ROOT 


325 


surface,  (Fig.  IV,)  is  composed  of  1  face  of  eacli  of  the  flat  blocks 
already  used,  2  faces  of  each  of  the  oblong  blocks,  and  3  faces  of 
the  little  cube.  But  we  have  in  the  complete  divisor,  15556,  1  face 
of  each  of  the  flat  blocks,  oblong  blocks,  and  little  cube  ;  and  in  the 
correction  of  the  trial  divisor,  856,  1  face  of  each  of  the  oblong  blocks 
and  of  the  little  cube  ;  and  in  the  square  of  the  last  root  figure,  16,  a 
third  face  of  the  little  cube.  Hence,  16  +  856  +  15556  =  16428,  the 
sisrnificant  fierures  of  the  new  trial  divisor.     This  number  is  obtained 


I. 

OPKB  ATION — CONTINUED. 

413494  1  74.5 
11.                       343 

in  the  opera- 
tion by  adding 
the  square  of 
the    last    root 

214 

1  14700 
856      15556 

70494 
62224 

figure  mental- 
ly, and  com- 
bining units 
of  like  order, 
thus:     16,    6, 

222.5 

111.25 

1642800 
16539.25 

8270.000 
8269.625 

.375 

and  6  are  28, 
and  we  write 

the  unit  figure  in  the  new  trial  divisor ;  then  2  to  carry,  and  5  and  5 
are  12,  etc.  We  annex  2  ciphers  to  this  trial  divisor,  as  to  the  former, 
and  dividing,  obtain  5,  the  third  figure  in  the  root.  To  complete  the 
second  trial  divisor,  after  the  manner  of  the  first,  the  correction  may 
be  found  by  annexing  .5  to  3  times  the  former  figures,  74,  and  multi- 
plying this  number  by  .5.  But  as  we  have,  in  column  I,  3  times  7, 
with  4  annexed,  or  214,  we  need  only  multiply  the  last  figure,  4,  by  3,  and 
annex  ,5,  making  222.5,  which  multiplied  by  .5  gives  111.25,  the  correc- 
tion required.  Then  we  obtain  the  complete  divisor,  16539.25,  the  pro- 
duct,  8269.625,  and  the  remainder,  .375,  in  the  manner  shown  by  the  for- 
mer steps.     From  this  example  and  analysis  we  deduce  the  following 

EuLE.  I.  Point  off  the  given  number  into  periods  of  three 
figures  each,  counting  from  units''  place  toward  the  left  and 
right. 

II.  Find  the  greatest  cube  that  does  not  exceed  the  left  hand 
2)eriod,  and  write  its  root  for  the  first  figure  in  the  required 
root ;  subtract  the  cube  from  the  left  hand  period,  and  to  the 
remainder  bring  down  the  next  period  for  a  dividend. 

III.  At  the  left  of  the  dividend  write  three  times  the  square 
of  the  first  figure  of  the  root,  and  annex  two  ciphers,  for  a  trial 
divisor;  divide  the  dividend  by  the  trial  divisor,  and  write 
the  quotient  for  a  trial  figure  in  the  root. 


326 


EVOLUTIOK. 


IV.  Annex  tlie  trial  figure  to  three  times  the  former  figure, 
and  write  the  result  in  a  column  marked  I,  one  line  below  the 
trial  divisor  ;  multiply  this  term  hy  the  trial  figure,  and  write 
the  product  on  the  same  line  in  a  column  marked  11;  add 
this  term  as  a  correction  to  the  trial  divisor,  and  the  result 
ivill  he  the  complete  divisor. 

V.  Multiply  the  complete  divisor  hy  the  trial  figure,  and 
suhtract  the  product  from  the  dividend,  and  to  the  remainder 
hring  down  the  next  period  for  a  new  dividend. 

VI.  Add  the  square  of  the  last  figure  of  the  root,  the  last 
term  in  column  II,  and  the  complete  divisor  together,  and  an- 
nex two  ciphers,  for  a  new  trial  divisor  j  with  which  obtain 
another  trial  figure  in  the  root. 

VII.  Multiply  the  unit  figure  of  the  last  term  in  column  I 
hy  3,  and  annex  the  trial  figure  of  the  root  for  the  next  term 
of  column  1 ;  multiply  this  result  hy  the  trial  figure  of  the  root 
for  the  next  term  of  column  II ;  add  this  term  to  the  trial 
divisor  for  a  complete  divisor,  with  which  proceed  as  before. 

1.  If  at  any  time  the  product  be  greater  than  the  dividend,  diminish  the  trial 
figure  of  the  root,  and  correct  the  erroneous  work. 

2.  If  a  cipher  occur  in  the  root,  annex  two  more  ciphers  to  the  trial  divisor,  and 
another  period  to  the  dividend ;  then  proceed  as  before  with  column  I,  annexing 
both  cipher  and  trial  figure. 

Examples  for  Practice. 
1.  What  is  the  cube  root  of  79.112  ? 

OPERATION. 

79.112  I  4.2928  +  ,  Ans. 
64. 


122 

244 

4800 
5044 

15112 
10088 

1269 

11421 

529200 
540621 

5024000 
4865589 

12872 

25744 

55212300 
55238044 

158411000 
11047G088 

128768 

1030144 

5526379200  47934912000 
5527409344  44219274752 

3715637248 

CUBE    ROOT.  327 

2.  What  is  the  cube  root  of  84604519  ?  Ans.  439. 

3.  What  is  the  cube  root  of  2357947691  ?  Ans.  1331. 

4.  What  is  the  cube  root  of  10963240788375  ?  Ans,  22215. 

5.  What  is  the  cube  root  of  270671777032189896  ? 

Ans.  646866. 

6.  What  is  the  cube  root  of  .091125  ?  Ans,   .45. 

7.  What  is  the  cube  root  of  .000529475129  ?    Ans,  .0809. 

8.  What  is  the  approximate  cube  root  of  .008649  ? 

Ans.  .2052 -f. 
Extract  the  cube  roots  of  the  foUowinof  numbers : 


-^2  =  1.259921  + 
^3  =  1.442249 -h 

^4  =  1.587401  + 


^5=  1.709975  + 
^6  =  1.817120  + 
^7  =  1.912931  + 


Applications  ik  Cube  Root. 

1.  What  is  the  length  of  one  side  of  a  cistern  of  cubical 
form,  containing  1331  solid  feet  ?  Ans.  11  feet. 

2.  The  pedestal  of  a  certain  monument  is  a  square  block 
of  granite,  containing  373248  solid  inches;  what  is  the 
length  of  one  of  its  sides  ?  Ans.  6  feet. 

3.  A  cubical  box  contains  474552  solid  inches ;  what  is 
the  area  of  one  of  its  sides  ?  Ans.  42^  sq.  ft. 

4.  How  much  paper  will  be  required  to  make  a  cubical 
box  which  shall  contain  ^ot  a,  solid  foot  ?    Ans.  I  of  sl  yard. 

5.  A  man  wishes  to  make  a  bin  to  contain  125  bushels,  of 
equal  width  and  depth,  and  length  double  the  width ;  what 
must  be  its  dimensions?  Ans,  Width  and  depth,  51.223  + 
inches  ;  length,  102. 446  +  inches. 

Spheres  are  to  each  other  as  the  cobes  of  their  diameters  or  circumferences. 

6.  There  are  two  spheres  whose  solid  contents  are  to  each 
other  as  27  to  343  ;  what  is  the  ratio  of  their  diameters  ? 

Analysis.  Since  spheres  are  to  each  other  as  the  cubes  of  their 
diameters,  the  diameters  will  be  to  each  other  as  the  cube  roots  of  the 
spheres  ;  and  «y/27  =  3,  -y/S^S  =  7  ;  hence  the  diameters  required  are 
as  3  to  7. 


328  ARITHMETICAL     PROGRESSION^. 

7.  The  diameter  of  a  sphere  containing  1  solid  foot  is  14.9 
inches ;  what  is  the  diameter  of  a  sphere  containing  2  solid 
feet  ?  Ans.  18.7  +  inches. 

8.  If  a  cable  4  in.  in  circumference,  will  support  a  sphere 
2  ft.  in  diameter,  what  is  the  diameter  of  that  sphere  which  will 
be  supported  by  a  cable  5  in.  in  circumference  ?  A71S.  2.32  +  ft. 

ARITHMETICAL    PROGRESSION. 

445.  An  Arithmetical  Progression,  or  Series,  is  a 

series  of  numbers  increasing  or  decreasing  by  a  common  dif- 
ference. Thus,  3,  5,  7,  9,  11,  etc.,  is  an  arithmetical  pro- 
gression with  an  ascending  series,  and  13, 10,  7,  4,  etc.,  is  an 
arithmetical  progression  with  a  descending  series. 

446.  The  Terms  of  a  series  are  the  numbers  of  which 
it  is  composed. 

447.  The  Extremes  are  the  first  and  last  terms. 

448.  The  Means  are  the  intermediate  terms. 

449.  The  Common  I>ifference  is  the  difference  between 
any  two  adjacent  terms. 

450.  There  are  five  parts  in  an  arithmetical  series,  any 
three  of  which  being  given,  the  other  ttoo  may  be  found. 
They  are  as  follows :  the  Ji7'st  term,  last  term,  common  dif- 
ference, number  of  terms,  and  sum  of  all  the  terms. 

Case  I. 

451 .  To  find  the  last  term  when  the  first  term, 
common  difierence,  and  number  of  terms  are  given. 

Let  2  be  the  first  term  of  an  ascending  seiies,  and  3  the 
common  difference ;  then  the  series  will  be  written,  2,  5,  8, 
11, 14,  or  analyzed  thus  :  2,  2  +  3,2  +  3  +  3,  2-f-3  +  3  + 
3,  2  +  3  +  3  +  3  +  3. 

Here  we  see  that,  in  an  ascending  series,  we  obtain  the 
second  term  hy  adding  the  common  difference  once  to  the  first 
term ;  the  third  term,  by  adding  the  common  difference 
twice  to  the  first  term ;  and,  in  general,  we  obtain  a7iy  term 


ARITHMETICAL    PROGRESSION.  329 

by  adding  the  common  difference  as  many  times  to  the  first 

term  as  there  are  terms  less  one. 

% 

The  analysis  for  a  descending  series  would  be  similar. 

Rule.  Multiply  the  common  difference  hy  the  number  of 
terms  less  one,  and  add  the  product  to  the  first  term,  if  the 
series  he  ascending,  and  subtract  it  if  the  series  be  descending. 

Examples  for  Practice. 

1.  The  first  term  of  an  ascending  series  is  4,  the  common 
difference  3,  and  the  number  of  terms  19  ;  what  is  the  last 
term?  Ans.  58. 

2.  What  is  the  13th  term  of  a  descending  series  whose 
first  term  is  75,  and  common  difference  5  ?  Ans,  \6. 

3.  A  boy  bought  18  hens,  paying  2  cents  for  the  first,  5 
cents  for  the  second,  and  8  cents  for  the  third,  in  arith- 
metical progression ;  what  did  he  pay  for  the  last  hen  ? 

4.  What  is  the  40th  term  of  the  series  i,  f,  1,  IJ,  etc.  ? 

Ans.  lOJ. 

5.  A  man  travels  9  days  ;  the  first  day  he  goes  20  miles, 
the  second  25  miles,  increasing  5  miles  each  day ;  how  far 
does  he  travel  the  last  day  of  his  journey?      Ans,  60  miles. 

6.  What  is  the  amount  of  $100,  at  7  per  cent.,  for  45 
years  ?  $100  +  $7  x  45  =  $415,  Ans. 

Case  II. 

452.  To  find  the  common  difference  when  the 
extremes  and  number  of  terms  are  given. 

Referring  to  the  series,  2,  5,  8,  11,  14,  analyzed  in  451, 
we  readily  see  that,  by  subtracting  the  first  term  from  any 
term,  we  have  left  the  common  difference  taken  as  many 
times  as  there  are  terms  less  one  ;  thus,  by  taking  away  2  in 
the  fifth  term,  2  -{-  3  -f-  3  -{-  3  +  3,  we  have  3  taken  4  times. 

Rule.  Divide  the  difference  of  the  extremes  by  the  number 
of  terms  less  one. 


330  akithmetical  progression. 

Examples  for  Practice. 

1.  The  first  term  is  2,  the  last  term  is  17,  and  the  number 
of  terms  is  6;  what  is  the  common  difference?      Ans,  3. 

2.  A  man  has  seven  children,  whose  ages  are  in  arith- 
metical progression ;  the  youngest  is  2  years  old,  and  the 
eldest  14 ;  what  is  the  common  difference  of  their  ages  ? 

Ans.  2  years. 

3.  The  extremes  of  an  arithmetical  series  are  1  and  50J,  and 
the  number  of  terms  is  34 ;  what  is  the  common  difference  ? 

4.  An  invalid  commenced  to  walk  for  exercise,  increasing 
the  distance  daily  by  a  common  difference  ;  the  first  day  he 
walked  3  miles,  and  the  14th  day  9^  miles ;  how  many  miles 
did  he  walk  each  day  ? 

When  we  have  found  the  common  difference  we  may  add  it  once,  twice,  etc,  to 
the  first  term,  and  we  have  the  series,  and  consequently  the  means. 

Ans.  3,  3^,  4,  4|,  5,  5^,  etc. 

Case  III. 

453.  To  find  the  number  of  terms  when  the  ex- 
tremes and  common  difference  are  given. 

Examining  the  series,  2,  5,  8,  11,  14,  analyzed  in  451, 
we  also  see  that  after  taking  away  the  Jirst  term  from  any 
term,  we  have  left  the  common  difference  taken  as  many 
times  as  the  number  of  terms,  less  1. 

Eule.  Divide  the  difference  of  the  extremes  ly  the  common 
difference,  and  add  1  to  the  quotient. 

Examples  for  Practice. 

1.  The  extremes  are  7  and  43,  and  the  common  difference 
is  4  ;  what  is  the  number  of  terms  ?  Ans,  10. 

2.  The  first  term  is  2J,  the  last  term  is  40,  and  the  common 
difference  is  7^ ;  what  is  the  number  of  terms  ?     Ans,  6. 

3.  A  laborer  agreed  to  build  a  fence  on  the  following  con- 
ditions :  for  the  first  rod  he  was  to  have  6  cents,  with  an  in- 
crease of  4  cents  on  each  successive  rod  ;  the  last  rod  came 
to  226  cents ;  how  many  rods  did  he  build  ?    Ans.  56  rods. 


GEOMETRICAL    PROGRESSION.  331 

Case  IV. 

454.  To  find  the  sum  of  all  the  terms  when  the 
extremes  and  number  of  terms  are  given. 

To  deduce  a  rule  for  finding  the  sum  of  all  the  terms,  we 
will  take  the  series  2,  5,  8,  11,  14,  writing  it  under  itself  in 
an  inverse  order,  and  add  each  term  ;  thus, 

2 -f-    5  +    8  H- 11  +  14  =  40,  once  the  sum. 
14  +  11  +    8  -f-    5  +    2  =  40,      "     ''      '' 
16  +  16  +  IG  +  16  +  16  =  80,  twice  the  sum. 
Here  we  perceive  that  16,  the  sum  of  the  extremes,  mul- 
tiplied by  5,  the  number  of  terms,  equals  80,  which  is  twice 
the  sum  of  the  series.    Dividing  80  by  2  gives  40,  which  is 
the  sum  required. 

Rule.  Multiply  the  sum  of  the  extremes  dy  the  number  of 
terms,  and  divide  the  product  by  2. 

Examples  for  Practice. 

1.  The  extremes  are  5  and  32,  and  the  number  of  terms 
12  ;  what  is  the  sum  of  all  the  terms  ?  Ans,  222. 

2.  How  many  strokes  does  a  common  clock  make  in  12 
hours?  Ans.  78  strokes. 

3.  What  debt  can  be  discharged  in  a  year  by  weekly  pay- 
ments in  arithmetical  progression,  the  first  being  $24,  and 
the  last  $1224  ?  Ans.  132448. 

4.  Suppose  100  apples  were  placed  in  a  line  2  yards  apart, 
and  a  basket  2  yards  from  the  first  apple  ;  how  far  would  a 
boy  travel  to  gather  them  up  singly,  and  return  with  each 
separately  to  the  basket  ?  Ans.  20200  yards. 


GEOMETEIOAL   PEOGRESSIOH". 

455*  A  Geometrical  Progression  is  a  series  of  num- 
bers increasing  or  decreasing  by  a  constant  multiplier. 
When  the  multiplier  is  greater  than  a  unit,  the  series  is 


GEOMETKICAL    PROGRESSION. 

ascending ;  thus,  2,  6,  18,  54,  162,  is  an  ascending  series, 
in  which  3  is  the  multiplier. 

When  the  multiplier  is  less  than  a  unit,  the  series  is  de- 
scending ;  thus,  162,  54,  18,  6,  2,  is  a  descending  series,  in 
which  -J-  is  the  multiplier. 

456.  The  Ratio  is  the  constant  multiplier. 

45'7«  In  every  geometrical  progression  there  are  five 
parts  to  be  considered,  any  three  of  which  being  given,  the 
other  two  may  be  determined.  They  are  as  follows :  The 
first  term,  last  term,  ratio,  number  of  terms,  and  the  sum  of 
all  the  terms. 

The  first  and  last  terms  are  the  extremes,  and  the  inter- 
mediate terms  are  the  means. 

Case  I. 

458.  To  find  any  term,  the  first  term,  the  ratio, 
and  number  of  terms  being  given. 

The  first  term  is  supposed  to  exist  independently  of  the 
ratio.  Using  the  ratio  once  as  a  factor,  we  have  the  second 
term ;  using  it  twice,  or  its  second  power,  we  have  the  third 
term ;  using  it  three  times,  or  its  third  power,  we  have  the 
fourth  term  ;  and,  in  general,  the  power  of  the  ratio  in  any 
term  is  one  less  than  the  number  of  the  term.  The  ascend- 
ing series,  2,  6,  18,  54,  may  be  analyzed  thus :  2,  2  x  3,  2 
x3x  3,  2x3x3x3. 

In  this  illustration  we  see  that 

1st  term,  2,  is  independent  of  the  ratio. 

2d  term,  6  =  2x3  =  the  first  term  into  the  1st  power  of 
the  ratio. 

3d  term,  18  =  2  x  3®  =  the  first  term  into  the  2d  power 
of  the  ratio. 

4th  term,  54  =  2  x  3^  =  the  first  term  into  the  3d  power 
of  the  ratio. 

Rule.  Multiply  the  first  term  by  that  power  of  the  ratio 
denoted  by  the  number  of  terms  less  1. 


geometrical  progression.  333 

Examples  for  Practice. 

1.  The  first  term  of  a  geometrical  series  is  4,  the  ratio  is 
3  ;  what  is  the  9th  term?  Ans.  4  x  3^  =  26244. 

2.  The  first  term  is  1024,  the  ratio  |-,  and  the  number  of 
terms  8 ;  what  is  the  last  term  ?  Ans.  ^. 

3.  A  boy  bought  9  oranges,  agreeing  to  pay  1  mill  for 
the  first  orange,  2  mills  for  the  second,  and  so  on ;  what 
did  the  last  orange  cost  him  ?  A7is.  $.256. 

4.  The  first  term  is  7,  the  ratio  ^,  and  the  number  of 
terms  7  ;  what  is  the  last  term  ?  A?is.  ts-J-oT' 

6.  What  is  the  amount  of  $1  at  compound  interest  for  5 
years,  at  7  per  cent,  per  annum  ?  Ans.  S1.40255  +  . 

In  the  above  example  the  first  term  is  $1,  the  ratio  is  $1.07,  and  the  number  of 
terms  is  6. 

6.  A  drover  bought  7  oxen,  agreeing  to  pay  $3  for  the 
first  ox,  $9  for  the  second,  $27  for  the  third,  and  so  on  ; 
what  did  the  last  ox  cost  him?  Ans.  $2187. 

Case  II. 

459.  To  find  the  sum  of  all  the  terms,  the  ex- 
tremes and  ratio  being  given. 

If  we  take  the  series  2,  8,  32, 128,  512,  in  which  the  ratio 
is  4,  multiply  each  term  by  the  ratio,  and  add  the  terms 
thus  multiplied,  we  shall  have 

8  +  32  +  128  +  512  +  2048  =  2728  = 

But  2  +  8  +  32  +  128  +  512  = 682= 

Hence,  by  subtracting,  we  get  2048-2  =  2046  = 
Dividing  by  3,  the  ratio  less  one,  2046^3  =  682  = 


Four  times  the  sum  of 

all  the  terms. 
Once  the  sum  of  all  th« 

terms. 
Three  times  the  sum  of 

all  the  terms. 
Once  the  sum  of  aU  the 

terms. 


The  subtraction  is  performed  by  taking  the  lower  line  or 
series  from  the  upper.  All  the  terms  cancel  except  2048 
and  2.  Taking  their  difference,  which  is  3  times  the  sum, 
and  dividing  by  3,  the  ratio  less  one,  we  must  have  the 
sum  of  all  the  terms. 


334  PROMISCUOUS     EXAMPLES. 

Rule.  Multiply  the  greater  extreme  hy  the  ratio,  subtract 
the  less  extreme  from  the  product,  and  divide  the  remainder 
hy  the  ratio  less  1. 

Let  every  decreasing  series  be  inverted,  and  the  first  term  called  the  last ;  then  the 
ratio  will  be  greater  than  a  unit.    If  the  series  be  irifinile,  the  first  term  is  a  cipher. 

Examples  for  Practice. 

1.  The  first  term  is  2,  the  last  term  486,  and  the  ratio  3  ; 
what  is  the  sum  of  all  the  terms?  A7is.  728. 

2.  The  first  term  is  4,  the  last  term  is  262144,  and  the 
ratio  is  4  ;  what  is  the  sum  of  the  series  ?      A7is,  349524. 

3.  The  first  term  of  a  descending  series  is  162,  the  last 
term  2,  and  the  ratio  -J-;  what  is  the  sum  ?         Ans.  242. 

4.  What  is  the  value  of  ^,  -gig-,  yj^,  etc.,  to  infinity  ?  Ans.  J. 

In  the  following  examples  we  first  find  the  last  term  by  the  Rule  under  Case  I. 

5.  What  yearly  debt  can  be  discharged  by  monthly  pay- 
ments, the  first  being  $2,  the  second  $6,  and  the  third  $18, 
and  so  on,  in  geometrical  progression  ?        Ans.  $531440. 

6.  If  a  grain  of  wheat  produce  7  grains,  and  these  be 
sown  the  second  year,  each  yielding  the  same  increase,  how 
many  bushels  will  be  produced  at  this  rate  in  12  years,  if 
1000  grains  make  a  pint  ?  Aiis.  252315  bu.  4-J  qt. 

7.  Six  persons  of  the  Morse  family  came  to  this  country 
200  years  ago  ;  suppose  that  their  number  has  doubled  every 
20  years  since,  what  would  be  their  number  now? 

The  other  cases  in  Progression  will  be  found  in  the  Higher  Arithmetic. 


PROMISCUOUS    EXAMPLES. 

1.  One-half  the  sum  of  two  numbers  is  800,  and  one-half  the  differ- 
ence of  the  same  numbers  is  200  ;  what  are  the  numbers  ? 

Ans.  1000  and  600. 

3.  What  number  is  that  to  which,  if  you  add  f  of  ^  of  itself,  the 
sum  will  be  61  ?  Ana.  55. 

8.   What  part  of  a  day  is  3  h.  21  min.  15  sec?  Ann.  ^^Vj- 


PROMISCUOUS     EXAMPLES.  335 

4.  A  commission  merchant  received  70  bags  of  wheat,  each  contain- 
ing 3  bu.  3  pk.  3  qt. ;  how  many  bushels  did  he  receive? 

5.  Four  men,  A,  B,  C,  and  D,  are  in  possession  of  $1100  ;  A  has  a 
certain  sum,  B  has  twice  as  much  as  A,  C  has  $300,  and  D  has  $200 
more  than  C  ;  how  many  dollars  has  A  ?  Atis.  $100. 

6.  At  a  certain  election,  3000  votes  were  cast  for  three  candidates, 
A,  B,  and  C ;  B  had  200  more  votes  than  A,  and  C  had  800  more  than 
B ;  how  many  votes  were  cast  for  A  ?  Ans.  600. 

7.  What  part  of  17^  is  3^  ?  Ans.  jf . 

8.  The  difference  between  f  and  |  of  a  number  is  10  ;  what  is  the 
number?  A718.  560. 

9  A  merchant  bought  a  hogshead  of  rum  for  $28.35  ;  how  much 
water  must  be  added  to  reduce  the  fii-st  cost  to  35  cents  per  gallon  ? 

Ans.  18  gal. 

10.  A  and  B  traded  with  equal  sums  of  money ;  A  gained  a  sum 
equal  to  ^  of  his  stock  ;  B  lost  $200,  and  then  he  had  |  as  much  as  A  ; 
how  much  was  the  original  stock  of  each  ?  A71S.  $500. 

11.  A  farmer  sold  17  bushels  of  barley,  and  13  bushels  of  wheat,  for 
$31.55 ;  he  received  for  the  wheat  35  cents  a  bushel  more  than  for  the 
barley  ;  what  was  the  price  of  each  per  bushel  ? 

Ans.  Barley,  $.C0  ;  wheat,  $1.25. 
13.  What  is  the  interval  of  time  between  March  20,  21  minutes  past 
3  o'clock,  P.  M.,  and  April  11,  5  minutes  past  7  o'clock,  A.  m.  ? 

Ans.  21  da.  15  h.  44  min. 

13.  What  o'clock  is  it  when  the  time  from  noon  is  ^\  of  the  time  to 
midnight?  Ans.  5  o'clock  24  min.  p.  m. 

14.  What  is  the  least  number  of  gallons  of  wine  that  can  be  ship- 
ped in  either  hogsheads,  tierces,  or  barrels,  just  filling  the  vessels, 
without  deficit  or  excess  ?  Ans.  126  gal. 

15.  A  ferryman  has  four  boats  ;  one  will  carry  8  barrels,  another  9, 
another  15,  and  another  16  ;  what  is  the  smallest  number  of  barrels 
that  will  make  full  freight  for  any  one,  and  all  of  the  boats  ? 

16.  A  and  B  have  the  same  income;  A  saves  ^  of  his,  but  B,  by 
spending  $30  a  year  more  than  A,  at  the  end  of  four  years  finds  him- 
self $40  in  debt ;  what  is  their  income,  and  how  much  does  each  spend 
a  year  ?  ( Income,     $160. 

Ans.  <  A  spends  $140. 
(B  spends  $170. 

17.  If  a  load  of  plaster  weighing  1825  pounds  cost  $2.19,  how  much 
is  that  per  ton  of  2000  pounds  ?  Ans.  $2.40. 

18.  If  2 1-  yards  of  cloth  If  yards  wide  cost  $3.37f,  what  will  be  the 
cost  of  36|  yards  1|  yards  wide  ?  Ans.  $52,779. 

19.  I  lend  my  neighbor  $200  for  6  months ;  how  long  ought  he  to 
lend  me  $1000  to  balance  the  favor  ? 

20.  Bought  railroad  stock  to  the  amount  of  $2356.80,  and  found  that 
the  sum  invested  was  40  per  cent,  of  what  I  had  left ;  what  sum  had 
I  at  first?  Ans.  $8248.80. 

21.  20  per  cent,  of  f  of  a  number  is  what  per  cent,  of  |  of  it  ? 


PROMISCUOUS    EXAMPLES. 

22.  Divide  a  prize  of  $10200  among  60  privates,  6  subaltern  officers, 
8  lieutenants,  and  a  commander,  giving  to  each  subaltern  double  the 
share  of  a  private,  each  lieutenant  8  times  as  much  as  the  subaltern, 
and  to  the  commander  double  that  of  a  lieutenant ;  how  much  is  each 
man's  share  ?  Ans.  Com,  $1200  ;  each  man,  $100. 

23.  A  is  51  miles  in  advance  of  B,  who  is  in  pursuit  of  him ;  A 
travels  16  miles  per  hour,  and  B  19  ;  in  how  many  hours  will  B  over- 
take A  ? 

24  How  much  wool,  at  20,  30,  and  54  cents  per  pound,  must  be 
mixed  with  95  pounds  at  50  cents,  to  make  the  whole  mixture  worth 
40  cents  per  pound  ? 

Ans.  183  lb.  at  20  ;  95  lb.  at  30  ;  190  lb.  at  54  cents. 

25.  If  240  bushels  of  wheat  are  purchased  at  the  rate  of  18  bushels 
for  $22|,  and  sold  at  the  rate  of  22 1  bushels  for  $33.75,  what  is  the 
profit  on  the  whole  ?  "  Ans.  $60. 

26.  My  horse,  wagon,  and  harness  together  are  worth  $169;  the 
wagon  is  worth  4  times  the  harness,  and  the  horse  is  worth  double 
the  wagon  ;  what  is  the  value  of  each  ?  i  Horse,      $104. 

Ans.  \  Wagon,   $  52. 
(  Harness,  $  13. 

27.  The  shadow  of  a  tree  measures  42  feet ;  a  staff  40  inches  in 
length  casts  a  shadow  18  inches  at  the  same  time  ;  what  is  the  height 
of  the  tree  ?  Ans.  93^  feet. 

28.  If  a  piece  of  land  40  rods  long  and  4  rods  wide  make  an  acre, 
how  wide  must  it  be  to  contain  the  same  if  it  be  but  25  rods  long  ? 

Ans.  6|  rods. 

29.  A,  B,  and  C  are  employed  to  do  a  piece  of  work  for  $26.45  ;  A 
and  B  together  are  supposed  to  do  |  of  it;  A  and  C  ^,  and  B  and  C 
^f ,  and  paid  proportionally  ;  how  much  must  each  receive  ? 

30.  If  12  ounces  of  wool  make  2\  yards  of  cloth  that  is  6  quarters 
wide,  how  many  pounds  of  wool  will  it  take  for  150  yards  of  cloth  4 
quarters  wide  ? 

31.  Six  persons,  A,  B,  C,  D,  E,  and  F,  are  to  share  among  them 
$6300  ;  A  is  to  have  \  of  it,  B  |,  C  |,  D  is  to  have  as  much  as  A  and 
C  together,  and  the  remainder  is  to  be  divided  between  E  and  F  in 
the  proportion  of  3  to  5  ;  how  much  does  each  one  receive  ? 

32.  What  is  the  amount  of  $200  for  8  years  at  6  per  cent.  comp«und 
interest?  Ans.  $318,709. 

33.  A  garrison,  consisting  of  360  men,  was  provisioned  for  6  months  ; 
but  at  the  end  of  5  months  they  dismissed  so  many  of  the  men  that 
the  remaining  provision  lasted  5  months  longer  ;  how  many  men  were 
sent  away. 

34.  A  certain  principal,  at  compound  interest  for  5  years,  at  6  per 
cent.,  will  amount  to  $669.113 ;  in  what  time  will  the  same  principal 
amount  to  the  same  sum,  at  6  per  cent,  simple  interest  ? 

Ans.  5  yr.  7  mo.  19.3  + da. 

35.  Paid  $148,350  for  9728  feet  of  pine  lumber ;  how  much  was  that 
per  thousand  ? 


PKOMISCUOUS    EXAMPLES.  337 

36.  Comparing  two  numbers,  483  was  found  to  be  their  least  com- 
mon multiple,  and  23  their  greatest  common  divisor  ;  what  is  the 
product  of  the  numbers  compared  ?  Ans.  11,109. 

37.  Eight  workmen,  laboring  7  hours  a  day  for  15  days,  were  able 
to  execute  ^  of  a  job  ;  in  how  many  days  can  they  complete  the 
residue,  by  working  9  hours  a  day,  if  4  workmen  are  added  to  their 
number?  Ans.  15|  days. 

38.  If  a  hall  36  feet  long  and  9  feet  wide  require  36  yards  of 
carpeting  1  yard  wide  to  cover  the  floor,  how  many  yards  IJ  yards 
wide  will  cover  a  floor  60  feet  long  and  27  feet  wide  ? 

Ans.  144  yards. 

39.  A,  B,  and  C  traded  in  company ;  A  put  in  $1  as  often  as  B  put 
in  $3,  and  B  put  in  $2  as  often  as  C  put  in  $5 ;  B's  money  was  in 
twice  as  long  as  C's,  and  A's  twice  as  long  as  B's  ;  they  gained  1 52.50 ; 
how  much  was  each  man's  share  of  the  gain?  (  As,  $12. 

Ans.  \  B's,  $18. 
(  C's,  $22.50. 

40.  A  and  B  found  a  watch  worth  $45,  and  agreed  to  divide  the 
value  of  it  in  the  ratio  of  f  to  f  ;  how  much  was  each  one's  share  ? 

Ans  i  ^20,  A's. 
^^^*-  I  $25,  B's. 

41.  A  man  received  $33.25  interest  on  a  sum  of  money,  loaned 
5  years  previous,  at  7  per  cent. ;  what  was  the  sum  lent  ? 

Ans.  $95. 

42.  The  diameter  of  a  ball  weighing  32  pounds  is  6  inches ;  what  is 
the  diameter  of  a  ball  weighing  4  pounds  ?  Ans.  3  inches. 

43.  Divide  $360  in  the  proportion  of  2,  3,  and  4, 

Ans.  $80,  $120,  $160. 

44.  If  by  working  6|  hours  a  day  a  man  can  accomplish  a  job  in 
12|  days,  how  many  days  will  be  required  if  he  work  S^  hours  per 
day  ?  '  Ans.  9^-q  days. 

45.  An  open  court  contains  40  square  yards  ;  how  many  stones, 
9  inches  square,  will  be  required  to  pave  it  ?  Ans.  640. 

46.  A  drover  paid  $76  for  calves  and  sheep,  paying  $3  for  calves, 
and  $2  for  sheep  ;  he  sold  {-  of  his  calves  and  |-  of  his  sheep  for  ^23, 
and  in  so  doing  lost  8  per  cent,  on  their  cost ;  how  many  of  each  did 
he  purchase  ?  Ans.   12  calves  ;  20  sheep. 

47.  If  a  cistern,  17|  feet  long,  10|^  broad,  and  13  deep,  hold  546 
barrels,  how  many  barrels  will  that  cistern  hold  that  is  16  feet  long, 
7  broad,  and  15  deep  ?  A7is.  384  bbls. 

48.  If  12  men,  working  9  hours  a  day  for  15f  days,  were  able  to 
execute  f  of  a  job,  how  many  men  may  be  withdrawn,  and  the  residue 
be  finished  in  15  days  more,  if  the  laborers  are  employed"  only  7  hours 
a  day  ?  *  An^.  4  men. 

49.  A  general  formed  his  men  into  a  square,  that  is,  an  equal 
number  in  rank  and  file,  and  found  that  he  had  59  men  over  ;  and 
increasing  the  number  in  both  rank  and  file  by  1  man,  he  wanted  84 
men  to  complete  the  square;  how  many  men  had  he? 

Ans.  5100. 

11.P.  15 


338  PROMISCUOUS    EXAMPLES. 

50.  Bought  wheat  at  $1.50  per  bushel,  com  at  $.75  per  bushel,  and 
barley  at  $.60  per  bushel  ;  the  wheat  cost  twice  as  much  as  the  corn, 
and  the  corn  twice  as  much  as  the  barley  ;  of  the  sum  paid,  $243 
and  I"  of  the  whole  was  for  wheat,  and  $153  and  ^^  of  the  whole 
was  for  the  corn  ;  how  many  bushels  of  grain  did  I  purchase  ? 

Ans.  756. 

51.  Divide  $630  among  3  persons,  so  that  the  second  shall  have  |  as 
much  as  the  first,  and  the  third  |  as  much  as  the  other  two ;  what  is 
the  share  of  each  ?  ( 1st,  $240. 

A718.  \  2d,   $180. 
(3d,   $210. 

52.  Bought  a  hogshead  of  molasses  for  $28,  and  7  gallons  leaked 
out ;  at  what  rate  per  gallon  must  the  remainder  be  sold  to  gain 
20%? 

58.  20  per  cent,  of  f  of  a  number  is  how  many  per  cent,  of  2  times 
I  of  1^  times  the  number  ?  Ana.  7^. 

54.  B  and  C,  trading  together,  find  their  stock  to  be  worth  $3500, 
of  which  C  owns  $2100  ;  they  have  gained  40  per  cent,  on  their  first 
capital ;  what  did  each  put  in?  ^^^   j  B,  $1000. 

( C,  $lo00. 

55.  If  the  ridge  of  a  building  be  8  feet  above  the  beams,  and  the 
building  be  32  feet  wide,  what  must  be  the  length  of  rafters  ? 

56.  If  12  workmen,  in  13  days,  working  12  hours  a  day,  can  make 
up  75  yards  of  cloth,  f  of  a  yard  wide,  into  articles  of  clothing  ;  how 
many  yards,  1  yard  wide,  can  be  made  up  into  like  articles,  by  10  men, 
working  9  days,  8  hours  each  day  ?  Ans.  23x''^. 

57.  A  grocer  sells  a  farmer  100  pounds  of  sugar,  at  12  cents  a  pound, 
and  makes  a  profit  of  9  per  cent.  ;  the  farmer  sells  him  100  pounds  of 
beef,  at  6  cents  a  pound,  and  makes  a  profit  of  10  per  cent. ;  who  gains 
the  more  by  the  trade,  and  how  much  ? 

Ans.  The  grocer  gains  $.445  + more. 

58.  In  1  yr.  4  mo.  $311.50  amounted  to  $336.42,  at  simple  interest; 
what  was  the  rate  %  ?  An^.  Q. 

59.  Three  persons  engage  to  do  a  piece  of  work  for  $20  ;  A  and  B 
estimate  that  they  do  |  of  it,  A  and  C  that  they  do  f  of  it.  and  B 
and  C  that  they  do  f  of  it ;  according  to  this  estimate,  what  part  of 
the  $20  should  each  man  receive  ? 

Ans.  A's,  $llf  ;  B's,  $5^  ;  C's,  $2«. 

60.  Paid  $375,  at  the  rate  of  2\  per  cent.,  for  insurance  on  a 
cotton  factory  and  the  machinery  ;  for  what  amount  was  the  policj 
given  ? 

61.  A  merchant  bought  goods  in  Boston  to  the  amount  of  $1000, 
and  gave  his  note,  dated  Jan.  1,  1857,  on  interest  after  3  months  ; 
six  months  after  the  note  was  given  he  paid  $560,  and  5  months  after 
the  first  payment  he  paid  $406 ;  what  was  due  Aug.  23,  1859  ? 

Ana.  $66.63  +  . 

62.  If  f  of  A's  money  be  equal  to  .|  of  B's,  and  5  of  B's  be  equal 
to  I  of  C's,  and  f  of  C's  be  equal  to  |  of  D's,  and  D  has  $45  more  than 
C,  how  much  has  each  ?  j«.  i  A,  $378  ;  C,  $360. 

^^-  ^  B.  $336 ;  D,  $405. 


PROMISCUOUS    EXAMPLES.  339 

63.  A  owed  B  $900,  to  be  paid  in  3  years  ;  but  at  tlie  expiration  of 
9  months  A  agreed  to  pay  $300  if  B  would  wait  long  enough  for  the 
balance  to  compensate  for  the  advance  ;  how  long  should  B  wait  after 
the  expiration  of  the  3  years  ?  Ans.  Id^  mo. 

64.  A  certain  clerk  receives  $800  a  year  ;  his  expenses  equal  ^  ot 
what  he  saves  ;  how  mucn  of  his  salary  does  he  save  yearly  ? 

65.  A  merchant  sold  cloth  at  $1  per  yard,  and  made  10  per  cent 
profit  ;  what  would  have  been  his  gain  or  loss  had  he  sold  it  at  $.87^ 
per  yard?  Atis.  Loss,  3f  %. 


66.  What  is  the  cube  of  —^  Ans.  f|. 

67.  What  is  the  cube  root  of  :rTr7T  Ans.  #. 

149^  ^ 

68.  A  miller  is  required  to  grind  100  bushels  of  provender  worth  50 
cents  a  bushel,  from  oats  worth  20  cents,  corn  worth  35  cents,  rye 
worth  60  cents,  and  wheat  worth  70  cents  per  bushel ;  how  many 
bushels  of  each  may  he  take  ? 

69.  A  man  owes  $6480  to  his  creditors  ;  his  debts  are  in  arith- 
metical progression,  the  least  being  $40,  and  the  greatest  $500 ; 
required  the  number  of  creditors  and  the  common  difference  between 
the  debts.  a        j  24  creditors. 

^'^*-   "I  $20  difference. 

70.  Two  ships  sail  from  the  same  port ;  one  goes  due  north  128 
miles,  and  the  other  due  east  72  miles ;  how  far  are  the  ships  from 
each  other?  ^l/is.  146.86  + miles. 

71.  If  10  pounds  of  cheese  be  equal  in  value  to  7  pounds  of  butter, 
and  11  pounds  of  butter  to  2  bushels  of  corn,  and  14  bushels  of  corn 
to  8  bushels  of  rye,  and  4  bushels  of  rye  to  1  cord  of  wood  ;  how 
many  pounds  of  cheese  are  equal  in  value  to  10  cords  of  wood  ? 

A71S.  550. 

72.  A  and  B  traded  until  they  gained  6  per  cent,  on  their  stock  ; 
then  f  of  A's  gain  was  $18  ;  if  A's  stock  was  to  B's  as  f  to  i,  how 
much  did  each  gain,  and  what  was  the  original  stock  of  each  ? 

.  ,^     j  A's  gain,  $45  ;      stock,  $750. 
^"*'    iB's    "      $37.50;     "      $625. 

73.  If  20  men,  in  21  days,  by  working  10  hours  a  day,  can  dig  a 
trench  30  ft.  long,  15  ft.  wide,  and  12  ft.  deep,  when  the  ground  is 
called. 3  degrees  of  hardness,  how  many  men,  in  25  days,  by  working 
8  hours  a  day,  can  dig  another  trench  45  ft.  long,  16  ft.  wide,  and  18 
ft.  deep,  when  the  ground  is  estimated  at  5  degrees  of  hardness  ? 

Ans.  84. 

74.  Wishing  to  know  the  height  of  a  certain  steeple,  I  measured 
the  shadow  of  the  same  on  a  horizontal  plane,  27^  feet ;  I  then  erected 
a  10  feet  pole  on  the  same  plane,  and  it  cast  a  shadow  of  21  feet ; 
what  was  the  height  of  the  steeple  ?  Ans.  103|  ft. 

75.  A  can  do  a  piece  of  work  in  3  days,  B  can  do  3  times  as  much  in 
8  days,  and  C  5  times  as  much  in  12  days  ;  in  what  time  can  they  all 
do  the  first  piece  of  work  ?  Ans.   I  da. 


340  PROMISCUOUS    EXAMPLES. 

76.  A  person  sold  two  farms  for  $1890  each  ;  for  one  he  received  25 
per  cent,  more  than  its  true  value,  and  for  the  other  25  per  cent,  less 
than  its  true  value  ;  did  he  gain  or  lose  by  the  sale,  and  how  much  ? 

A718.   Lost  $252. 

77.  Three  men  paid  $100  for  a  pasture  ;  A  put  in  9  horses,  B  12 
cows  for  twice  the  time,  and  C  some  sheep  for  2^  times  as  long  as  B's 
cows ;  C  paid  one  half  the  cost ;  how  many  sheep  had  he,  and  how 
much  did  A  and  B  each  pay,  provided  6  cows  eat  as  much  as  4  horses, 
and  10  sheep  as  much  as  3  cows  ?  (  C  had  25  sheep. 

Ans.  <  A  paid  $18. 
(B.    "    $32. 

78.  A  man  purchased  goods  for  $10500,  to  be  paid  in  three  equal 
installments,  without  interest ;  the  first  in  3  months,  the  second  in  4 
months,  the  third  in  8  mc  nths  ;  how  much  ready  money  will  pay  the 
debt,  money  being  worth  7fo  ?  Ans.   $10203.94  +  . 

79.  A  farmer  sold  50  fowls,  consisting  of  geese  and  turkeys  ;  for  the 
geese  he  received  $.75  apiece,  and  for  the  turkeys  $1.25  apiece,  and 
for  the  whole  he  received  $52.50  ;   how  many  were  there  of  each  ? 

Ans.  20  geese,  30  turkeys. 

80.  There  is  an  island  73  miles  in  circumference,  and  3  footmen 
start  together  and  travel  around  it  in  the  same  direction ;  A  goes  5 
miles  an  hour,  B  8,  and  C  10;  in  what  time  will  they  all  come 
together  again  if  they  travel  12  hours  a  day  ?  Ans.   6  da.  1  h. 

81.  A,  B,  and  C  are  to  share  $100000  in  the  proportion  of  ^,  \-,  and  I, 
respectively ;  but  C  dying,  it  is  required  to  divide  the  whole  sum 
proportionally  between  the  other  two  ;  how  much  is  each  one's  share  ? 

An^    i  A's,  $57142.85f. 
^^'  I  B's,  $42857.14|. 

82.  A,  B,  and  C  have  135  sheep  ;  A's  plus  B's  are  to  B's  plus  C's 
as  5  to  7,  and  C's  minus  B's  to  C's  plus  B's  as  1  to  7  ;  how  many  has 
each  ?  Ans.   A,  £0  ;  B,  45  ;  C,  60. 

83.  A  man  sold  one  hog,  weighing  250  pounds,  at  4  cents  per 
pound  ;  a  second,  weighing  300  pounds,  at  4|  cents ;  and  a  third, 
weighing  369  pounds,  at  5  cents ;  what  was  the  average  price  per 
pound  for  the  whole  ?  Ans.  4^||  cents. 

84  In  a  certain  factory  are  employed  men,  women,  and  boys  ; 
the  boys  receive  3  cents  an  hour,  the  women  4,  and  the  men  6 ;  the 
boys  work  8  hours  a  day,  the  women  9,  and  the  men  12  ;  the  boys 
receive  $5  as  often  as  the  women  $10,  and  for  every  $10  paid  to  the 
women,  $24  are  paid  to  the  men  ;  how  many  men,  women,  and  boys 
are  there,  the  whole  number  being  59  ? 

Ans.  24  men,  20  women,  15  boys. 

85.  A  fountain  has  4  receiving  pipes,  A,  B,  C,  and  D  ;  A.  B,  and  C 
will  fill  it  in  6  hours,  B,  C,  and  D  in  8  hours,  C,  D,  and  A  in  10  hours, 
and  D,  A,  and  B  in  12  hours ;  it  has  also  4  discharging  pipes,  W,  X, 
Y,  and  Z ;  W,  X,  and  Y  will  empty  it  in  6  hours,  X,  Y,  and  Z  in  5 
hours,  Y,  Z,  and  W  in  4  hours,  and  Z,  W,  and  X  in  3  Hours  ;  suppose 
the  pipes  all  open,  and  the  fountain  full,  in  what  time  would  it  be 
emptied?  Ana.   Qt^^-h. 


PROMISCUOUS    EXAMPLES.  341 

56.  How  many  building  lots,  each  75  feet  by  125  feet,  can  be  laid 
out  on  1  A.  46  P.  18^  sq.  yd.  ?  Ans.  6. 

87.  A  man  bought  a  house,  and  agreed  to  pay  for  it  $1  on  the  first 
day  of  January,  $2  on  the  first  day  of  February,  $4  on  the  first  day  of 
March,  and  so  on,  in  geometrical  progression,  through  the  year ;  what 
was  the  cost  of  the  house,  and  what  the  average  time  of  payment  ? 

Ans  i^^^^^- 

■  ( Average  time,  Nov.  1. 

88.  A  man  sold  a  rectangular  piece  of  ground,  measuring  44  chains 
32  links  long  by  36  chains  wide ;  how  many  acres  did  it  contain  ? 

Am.  159  A.  88.32  P. 

89.  What  number  is  that  which  being  increased  by  its  half,  its  third, 
and  18  more,  will  be  doubled  ?  Ans.  108. 

90.  A  merchant  has  200  lb.  of  tea,  worth  |.63|  per  pound,  which  he 
will  sell  at  $.56  per  pound,  provided  the  purchaser  will  pay  in  coffee 
at  22  cents,  which  is  worth  25  cents  per  pound ;  does  the  merchant 
gain  or  lose  by  the  sale  of  the  tea,  and  how  much  per  cent.  ? 

Ans.  Gained  lj\%. 

91.  A  man  owes  a  debt  to  be  paid  in  4  equal  installments  at  4,9, 13, 
and  20  months,  respectively  ;  discount  being  allowed  at  5  per  cent.,  he 
finds  that  $750  ready  money  will  pay  the  debt ;  how  much  did  he  owe  ? 

^715.  $78474  +  . 

92.  A  and  B  traded  upon  equal  capitals  ;  A  gained  a  sum  equal  to  I 
of  his  capital,  and  B  a  sum  equal  to  |^  of  his  ;  B's  gain  was  $500  less 
than  A's ;  what  was  the  capital  of  each?  Ans.  $4000. 

93.  I  purchase  goods  in  bills  as  follows  :  June  4, 1859,  $240.75 ;  Aug. 
9.  1859,  $137.25;  Aug.  29,  1859.  $65.64;  Sept.  4,  1859,  $230.36;  Nov. 
12,  1859,  $36.  If  the  merchant  agree  to  allow  credit  of  6  mo.  on  each 
bill,  when  may  I  settle  by  paying  the  whole  amount? 

A?i8,  Feb.  1,  1860. 

94.  A  young  man  inherited  a  fortune,  -\  of  which  he  spent  in  3 
months,  and  f  of  the  remainder  in  10  months,  when  he  had  only  $2524 
left ;  how  much  had  he  at  first?  Ans.  $5889.33  +  . 

95.  A  man  bought  a  piece  of  land  for  $3000,  agreeing  to  pay  7  per 
cent,  interest,  and  to  pay  principal  and  interest  in  5  equal  annual 
installments ;  how  much  was  the  annual  payment  ? 

Ans.  $731.67  +  . 

96.  I  have  three  notes  parable  as  follows :  one  for  $200,  due  Jan.  1, 
1859,  another  for  $350,  due* Sept.  1,  and  another  for  $500,  due  April  1, 
1860  ;  what  is  the  average  of  maturity  ?  Ans.  Oct  24,  1859. 

97.  A  man  held  three  notes,  the  first  for  $600,  due  July  7,  1859  ; 
the  second  for  |530,  due  Oct.  4, 1859;  and  the  third  for  $400,  duo 
Feb.  20,  1860 ;  he  made  an  equitable  exchange  of  these  with  a  specu- 
lator for  two  other  notes,  one  of  which  was  for  $730,  due  Nov.  lo, 
1859  ;  what  was  the  face  of  the  other,  and  when  due  ? 

.^     (  Face,  $800. 
^^•1  Due  Aug.  29, 1859. 


342  MENSUIIATION. 


MENSURATION  OF  LINES  AND  SUPERFICIES. 

460.  In  taking  the  measure  of  any  line,  surface,  or  solid,  we  are 
always  governed  by  some  denomination,  a  unit  of  which  is  called  the 
Unit  of  Measure.  Thus,  if  any  lineal  measure  be  estimated  in  feet, 
the  unit  of  measure  is  1  foot ;  if  in  inches,  the  unit  is  1  inch.  If  any 
superficial  measure  be  estimated  in  feet,  the  unit  of  measure  is  1 
square  foot ;  if  in  yards,  the  unit  is  1  square  yard. 

4:(>1.  If  any  solid  or  cubic  measure  be  estimated  in  feet,  the  unit  of 
measure  is  1  cubic  foot ;  if  in  yards,  the  unit  is  1  cubic  yard. 

462.  The  area  of  a  figure  is  its  superficial  contents,  or  the 
surface  included  within  any  given  lines, 
without  regard  to  thickness. 

463.  An  Oblique  Anj^Ie  is  an 

angle  greater  or  less  than  a  right  angle  ; 

thus,  ABC  and  C  B  D  are  oblique  angles.       ^ 

Case  I. 

464.  To  find  the  area  of  a  square  or  a  rectangle. 

465.  A  Square  is  a  figure  having  four  equal  sides  and  four  right 
angles. 

466.  A  Rectang'le  is  a  figure  having  four  right  angles,  and  its 
opposite  sides  equal. 

Rule.  Multiply  the  length  ly  the  breadth,  and  the  product  mil  be  the 
square  contents. 

Examples  for  Practice. 

1.  How  many  square  inches  in  a  board  3  feet  long  and  20  inches 
wide?  Ans.  720. 

2.  A  man  bought  a  farm  198  rods  long  and  150  rods  wide,  and 
agreed  to  give  $32  an  acre ;  how  much  did  the  farm  cost  him  ? 

A71S.  15940. 

3.  A  certain  rectangular  piece  of  land  measures  1000  links  by  100 ; 
how  many  acres  does  it  contain  ?  Ans.  1  A. 

Case  II. 

467.  To  find  the  area  of  a  rhombus  or  a  rhomboid. 

468.  A  Rhombus  is  a  figure  having  four  equal  sides  and  four 
oblique  angles. 

469.  A  Rhomboid  is  a  figure  having  its  opposite  sides  equal 
and  parallel,  and  its  angles  oblique. 

The  pquare,  rcctancrle,  rhombus,  and  rhom\)oid,  having  their  opposite  Bides  par- 
allel,  are  called  by  the  general  name,  parallelogram. 

It  is  proved  in  geometry  that  any  parallelogram  is  equal  to  a  rect- 
angle of  the  same  length  and  width. 


MEIS-SUKATION-.  343 

Rule.  MuXtvply  fhe  length  by  the  shortest  or  p&rpendiaular  distance 
between  two  opposite  sides. 

Examples  for  Practice. 

1.  A  meadow  in  the  form  of  a  rliomboid  is  20  chains  long,  and  the 
shortest  distance  between  its  longest  sides  is  12  chains ;  how  many 
days  of  10  hours  each  will  it  take  a  man  to  m^w  the  grass  on  this 
meadow,  at  the  rate  of  1  square  rod  a  minute  ?  Ans.  6  da.  4  h. 

2.  The  side  of  a  plat  in  the  form  of  a  rhombus  is  15  feet,  and  a  per- 
pendicular  drawn  from  one  oblique  angle  to  the  side  opposite,  will 
meet  this  side  9  feet  from  the  adjacent  angle  ;  what  is  the  area  of  the 
plat  ?  Ans.  180  sq.  ft. 

Case  III. 

470.  To  find  the  area  of  a  trapezoid. 

471.  A  Trapezoid  is  a  figure  having  A  V"] 
four  sides,  of  which  two  are  parallel.                                 /  j  \    j 

The  mean  length  of  a  trapezoid  is  one  /  ;  \  j 

half  the  sum  of  the  parallel  sides.  /     |  \  I 

Rule.  Multiply  one  half  th^  sum  of  the  parallel  sides  by  the  perpen- 
dicular distance  between  tJiem. 

Examples  for  Practice. 

1.  What  are  the  square  contents  of  a  board  12  feet  long,  16  inches 
wide  at  one  end,  and  9  at  the  other?  Ans.  VZ^  sq.  ft. 

2.  What  is  the  area  of  a  board  8  feet  long,  16  inches  wide  at  each 
end,  and  8  in  the  middle?  Ans.  8  sq.  ft. 

3.  One  side  of  a  field  is  40  chains  long,  the  side  parallel  to  it  is 
22  chains,  and  the  perpendicular  distance  between  these  two  sides  is 
25  chains ;  how  many  acres  in  the  field?  An^.  Ill  A.  5  sq.  ch. 

Case  IV. 

472.  To  find  the  area  of  a  triangle. 

473.  The  Base  of  a  triangle  is  the  side  on  which  it  is  supposed 
to  stand. 

474.  The  Altitude  of  a  triangle  is  the  perpendicular  distance 
from  the  angle  opposite  the  base  to  the  base,  or  to  the  base  produced 
or  extended. 

475.  A  Triangrle  is  one  half  of  a  parallelogram  of  the  same 
base  and  altitude. 

Rule.  Multiply  one-half  the  base  by  the  altitude,  or  one-half  the 
altitude  by  the  base.  Or,  Multiply  the  base  by  the  altitude,  and  divide 
the  product  by  2. 

Examples  for  Practice. 

1.  How  many  square  yards  in  a  triangle  whose  base  is  148  sq.  feet, 
and  perpendicular  45  feet  ?  Ans.  870  sq.  yds. 


344  MENSURATION". 

2.  The  gable  ends  of  a  barn  are  each  28  feet  wide,  and  the  perpen- 
dicular height  of  the  ridge  above  the  eaves  is  7  feet ;  how  many  feet 
of  boards  will  be  required  to  board  up  both  gables  ?     An%.  196  feet. 

Case  V. 

476.  To  find  the  circnmference  or  the  diameter  of 
a  circle. 

477.  A  Circle  is  a  figure  bounded  by  one 
laniform  curved  line. 

478.  The  Circumference  of  a  circle  is 
the  curved  line  bounding  it. 

479.  The  Diameter  of  a  circle  is  a  straight 
line  passing  through  the  center,  and  terminating 
in  the  circumference. 

It  is  proved  in  geometry  that  in  every  circle  the  ratio  between  the 
diameter  and  the  circumference  is  3.1416 +  . 

Rule.    I.    To  find  the  circumference. — Multiply  tJie  diameter  by 
8.1416. 
II.   To  find  the  diameter. — Multiply  the  cvrcumfer&nce  by  .3183. 

Examples  fob  Pkactice. 

1.  Wliat  length  of  tire  will  it  take  to  band  a  carriage  wheel  5  feet 
in  diameter  ?  Ans.  15  ft.  8. 4  +  in. 

2.  What  is  the  circumference  of  a  circular  lake  721  rods  in  diame- 
ter ?  Ans.  7  mi.  25  rds.  1.54  +  ft. 

3.  What  is  the  diameter  of  a  circle  33  yards  in  circumference  ? 

Arhs.  10.5+ yards. 

Case  YL 

480.  To  find  the  area  of  a  circle. 

From  the  principles  of  geometry  is  derived  the  following 

Rule.  I.  When  both  diameter  and  circumference  are  given ; — 
Multiply  the  diameter  by  the  circumference,  and  divide  tJie  product  by  4. 

II.  When  the  diameter  is  given  ;  —  Multiply  the  aqua/re  of  the  diame- 
ter by  .7854. 

III.  When  the  circumference  is  given ; —  Multiply  the  square  of  the 
circumference  by  .07958. 

Examples  for  Practice. 

1.  The  diameter  of  a  circle  is  113,  and  the  circumference  355  ;  what 
Is  the  area?  Ans.  10028.75. 

2.  What  is  the  diameter  of  a  circular  island  containing  1  square 
mile  of  land?  Ans.  1  mi.  41  rd.  1.4  + ft. 

3.  A  man  has  a  circular  garden  requiring  84  rods  of  fencing  to  in- 
close it ;  how  much  land  in  the  garden  ?  Afis.  3  A.  81.5+  P. 


METRIC     SYSTEM. 


345 


THE    METEIC    SYSTEM.* 


"■^ iiii.ml.i..nti.nl,...„„...l..... 


'10.764 


The  three  dimensions  of  this  cube  are,  1  Meter,  or 

10  Decimeters,  or  100  Centimeters,  in  length. 

ScAiiB  A  OF  THE  Exact  Size. 


1  Cu.  Centimeter, 
Exact  Size. 


481.   The  Metric  System  of  Weights  and  Measures 

has  now  received  the  sanction  of  law  among  more  than  half 
the  inhabitants  of  the  civilized  world.  Up  to  this  date  it 
has  been  adopted  in  Erance,  Germany,  Austria,  the  Nether- 
lands, Southern  Europe,  and  South  America,  and  has  been 
legalized  in  Great  Britain,  Germany,  and  the  United 
States. 

*  The  Metric  System,  as  it  was  presented  in  all  tlie  editions  of  this 
book,  printed  previous  to  1877,  was  useless,  because  the  symbols  and 
applications  did  not  correspond  with  present  usage. 

This  will  be  a  sufficient  reason  for  substituting  in  place  of  the 
former  matter,  a  condensed  and  practical  treatise  of  the  system, 
together  with  some  useful  miscellaneous  tables. 


340 


METRIC     SYSTEM 


4:8S5.  The  Metric  System  of  weights  and  measures  is  based 
upon  the  decimal  scale. 

483.  The  Meter  is  the  ha^e  of  the  system,  and  is  the  one  ten 
millionth  part  of  the  distance  on  the  earth's  surface  from  the  equator 
to  either  pole,  or  39.37079  inches. 

From  the  Meter  are  made  the  Ar  (are),  the  Liter  (leeter),  and  the 
Oram  ;  these  constitute  the  'primary  or  principal  units  of  the  system, 
from  which  all  the  others  are  derived. 

484.  The  Multiple  Uuits,  or  higher  denominations,  are 
named  by  prefixing  to  the  name  of  the  primary  units  the  Greek 
numerals,  Deka  (10),  Hecto  (100),  Kilo  (1000),  and  Myra  (10000). 

485.  The  Sub-Mllltiple  Units,  or  lower  denominations,  are 
named  by  prefixing  to  tlie  names  of  the  primary  units  the  Latin 
numerals,  Bed  {-^-q),  Genti  (y^^),  Milli  (tttW)- 

Hence,  it  is  apparent  from  the  name  of  a  unit  whether  it  is  greater 
or  less  than  the  standard  unit,  and  also  how  many  times. 

Meter  means  measure ;  and  the  three  principal  units  are  lengthy 
capacity  or  volume,  and  weight. 


TABLES    AUTHORIZED     BY    CONGRESS. 
486.    MEASUEES    OF    LENGTH. 


Metric  Denomhiations. 


Millimeter. . 
Centimeter. 
Decimeter . . 

Meter 

Dekameter. , 
Hektometer. 
Kilometer . . 
Myriameter. 


AbbreTia- 
tions. 

Values 

Equivalents  legalized 
by  Congress  in  denom- 
inations now  in  use. 

mm. 

.001  M. 

0.03937  in. 

cm, 

.01 

M. 

0.3937    " 

dm. 

.1 

M. 

3.937      " 

M. 

1 

M. 

89.37       '* 

Dm. 

10 

M. 

893.7 

Hm. 

100 

M. 

828  ft.  1  in. 

Km. 
Mm. 

1,000 
10,000 

M. 
M. 

3280  ft.  10  in. 

or  0.62187  milee. 
6.2137  miles. 

The  Meter,  like  our  yard,  is  the  unit  used  in  measuring  doths 
and  short  distances. 

The  Kilometer  is  commonly  used  for  measuring  long  distances,  and 
is  about  f  of  a  common  mile. 


METRIC     SYSTEM.  347 

The  Centimeter  and  Millimeter  are  used  by  mechanics  and  others 
for  minute  lengths.  The  Dekameter,  Hektometer,  and  Myriameter  are 
seldom  used. 

The  Square  and  Gvbic  Measures  are  only  the  squares  and  cubes  of 
the  measures  of  length,  as  shown  in  the  following  tables. 


SQUARE    MEASURE. 

Table. 

1  Centar,   ca.     =1  Sq.  Meter    =1550  sq.  in. 
100  Centars,   " 


-^^a  a     AT  .        »  =1  -4»*  =119.6034  sq.  yd. 

100  Sq.  Meters.  J  ^  "^ 

..  i^  f^\.      ^'  \  =1  Hektar(5a.)=2.471  acres. 

10,000  Sq.  Meters,  f  ^       ' 

The  Square  Meter  is  the  unit  for  measuring  ordinary  surfaces  of 
small  extent. 

The  Ar  is  the  unit  of  land  measure,  and  is  a  square  whose  side  is 
10  meters,  equal  to  a  square  dekameter,  or  119.6  square  yards. 


CUBIC    MEASURE. 
Table. 


0[  =: 


1  Cubic  Meter      {Cu.  M.y 
1000  Cu.  Decimeters  {cu.  dm)\  =1  Ster       =35.316  cu.  ft. 
10  Decisters  {ds.) 

10  Sters  {S.)  =1  Dekaster= 13.079  cu.  yd. 


The  Cubic  Meter  is  the  uni4i  for  measuring  ordinary  solids. 

The  Ster  is  the  unit  of  wood  measure,  and  is  equal  to  a  culm 
meter,  or  .2759  cord. 

A  Cuhic  Meter  of  water  weighs  1000  kilos,  or  a  metric  ton. 

The  scale  for  the  meter,  liter  sinAgram  is  uniformly  10  ;  but  in  the 
scale  for  the  square  meter,  100  units  of  any  order  make  one  unit  of 
the  next  higher ;  and  in  the  scale  for  the  cubic  meter,  1000  units  of 
any  order  make  one  of  the  next  higher. 


348 


MET  HI  C     SYSTEM. 


487.  MEASURES    OF    CAPACITY 


Metric  Names. 

Abbrevi- 
ations. 

No.  of  Liters. 

Dry  Measure. 

Liquid  or  Wine 
Measure. 

Milliliter 

ml. 

cl. 

dl. 

L. 
Dl. 
HI. 
Kl. 

.001 
.01 
.1 
1 
10 
100 
1000 

0.061    cu.  in. 
0.6102   "     " 
6.1023   "     " 
0.908  qt. 
9.08  qts. 
2  bu.  3.35  pks. 
1.308  cu  yds. 

0  27  fld  dr 

Centiliter 

Deciliter 

0.338  fld.  o7, 
0.845  gill. 
1.0567  qts. 
2.6417igal8. 
26.417^    " 
264.17  V  gals. 

Liter 

Dekaliter 

Hektoliter 

Kiloliter 

The  Liter  is  commonly  used  in  measuring  liquids  in  moderate  quan- 
tities ;  the  dekaliter,  for  large  quantities  and  for  fruits,  etc. 

The  Hektoliter  is  used  for  measuring  grain,  fruit,  roots,  etc.,  in  large 
quantities,  also  for  casks  of  wine. 


488.      MEASURES    OF    WEIGHT 


Metric  DENOjnNATioisrs  and  VAiiUKS. 

Weight  of  what 

quantity  of  water 

at  maximum. 

Equivalent  in  de- 
nominations now 

Names. 

Abbrevi- 
ations. 

No.  of  Grams. 

in  use. 

Avoirdupois 

weight. 

Milligram 

Centigram 

Decigram 

Grrain 

mg. 
eg. 

dg. 

G. 
Dg. 
Hg. 
Kg. 
Mg. 

Q. 

T. 

.001 

.01 

.1 

10 

100 

1,000 

10,000 

100,000 

1,000,000 

1  cu.  millim. 
10   " 

.1  cu.centim. 

1     "       " 
10     "      - 

1  deciliter. 

1  liter. 
10  liters. 

1  hektoDter. 

1  cu.  meter. 

0.0154  T.gr. 
0.1543    " 
1.5432    '• 
15.432     " 

Dekagram 

Hektogram 

Kilogram  or  Kilo. 

Myriagram 

Quintal 

Ton  or  Millier 

0.3527  A.oz. 
3.5274    " 
2.2046  lbs. 
22.046      " 
220.46 
3204.6 

The  Kilogram,  or  Kilo,  is  the  unit  of  common  weight  in  tmde  and 
is  a  trifle  less  than  3^  lb.  Avoirdupois. 

The  Ton  is  used  for  weighing  very  heavy  articles,  and  is  about 
204  lb.  more  than  a  common  ton.  The  other  denominations,  larger 
than  the  gram,  are  seldom  used. 


METRIC     SYSTEM 


349 


489.    Principal  Units  and  their  Equivalents. 


Meter. 


Ar 


Liter 


Gram 


1.  The  unit  of  Length. 

2.  The  lase  of   the   metric   system   is  one 

ten-millionth  part  of  a  quadrant  of  the 
earth's  meridian. 

3.  Equivalent,  39.3708  inches. 

1.  The  unit  of  Surface. 

2.  A  square  whose  side  is  10  meters. 

3.  Equivalent,  119.6  square  yards. 

1.  The  unit  of  Capacity  or  Volume. 

2.  A  vessel  whose  volume  is  equal  to  a  cube 

whose  edge  is  one-tenth  of  a  meter. 

3.  Equivalent,  .908  quart  dry  measure,  or 

1.0567  quarts  wine  measure. 
A  liter  of  water  weighs  about  2^  pounds. 

1.  The  U7iit  of  Weight. 

2.  The  weight  of  a  cube  of  pure  water  whose 

edge  is  .01  of  a  meter. 

3.  The  water  must  be  weighed  in  a  vacuum 

4°  C,  or  39.2°  F. 

4.  Equivalent,  15.432  grains. 


490.  Approximate  Values. 

When  great  accuracy  is  not  required,  it  may  often 
be  convenient  to  use  the  following  approximate  values,  which 
are  very  nearly  the  correct  values  : 


1  Decimeter  =     4    Inches. 

1  Cu.  Meter 

=       i  Cord. 

1  Meter          =  40    Inches. 

1  Liter 

=     1    Quart. 

5  Meters        =     1    Rod. 

1  Hektoliter 

=     2^  Bushels 

1  Kilometer  =       f  Mile. 

1  Gram 

=  15^  Grains. 

1  Sq.  Meter    =  10|  Sq.Feet. 

1  Kilogram 

=     ^  Pounds. 

1  Hektar        ==     2^  Acres. 

1  Metric  Ton 

=2204  Pounds. 

A  Myriameter  is  10  kilometers  ;  an  ar  is  1   square  dekameter  of 
land  ;  a  ster  is  a  cubic  meter  of  fire-wood. 


J3 

o 

^ 

=1 

&H 

Eh 

w 

6 

7 

0 

a 

a 

g 

'^ 

w 

w 

350  METRIC     SYSTEM. 

491.  METRIC  NOTATION  AND  NUMERATION. 

The  Metric  System  is  based  upon  the  AraUc  Notation, 
the  denominations  corresponding  with  the  orders  of  the 
Arabic  notation ;  and  hence  is  written  like  United  States 
Money,  the  lowest  denomination  at  the  right,  as  illustrated 
in  the  following 

Table. 

§      ^      .§      a      -§       I 

®      .^      J      .§      5      •« 
H      t^       Q      Eh      W       H 

15.638 

Is-      I  i  i 

The  number  is  read  67015.638  meters  ;  or  if  expressed  in 
centimeters  it  is  read  6701563.8  centimeters  ;  if  in  hilometers, 
it  is  read  67.015638  kilometers  ;  or,  it  may  be  read  6  myria- 
meters,  7  kilometers,  2  hektometers,  1  dekameter,  5  meters, 
6  decimeters,  3  centimeters,  8  millimeters. 

The  names  mill^  cent,  dime,  used  in  United  States  Money,  corres- 
pond to  milU,  centi,  deci,  in  the  Metric  System.  Hence  the  eagle 
might  be  called  the  deka-dollar,  since  it  is  10  dollars  ;  the  dime,  a  deci- 
dollar,  since  it  is  j*^  of  a  dollar,  etc. 

Write  3672.045  grams,  and  read  it  in  the  several  orders ; 
read  it  in  hektograms ;  in  kilograms ;  in  decigrams ;  in 
centigrams. 

Rule. —  Write  in  their  consecutive  order  the  several  detiom" 
mations,  and  place  a  decimal  point  between  the  given  denom^ 
ination  and  those  of  lower  denominations. 

1.  Read  1456.34  liters,  in  metric  units;  read  it  in  centi- 
liters; in  dekaliters  ;  in  hektoliters. 

2.  Read  795  centars  in  metric  units  ;  in  ars. 

3.  Write  6  kilometers,  7  dekameters,  and  3  decimeters,  as 
meters.  Ans.  6070.3  meters. 

4.  How  many  liters  in  5  kiloliters,  4  hektoliters,  7  liters. 
6  centiliters  ? 


METRIC     SYSTEM.  351 

492.  To  express  a  higher  Metric  Denomination 
in  terms  of  its  lower  denominations. 

1.  Reduce  34.6732  kilometers  to  meters. 
Operation.    34. 6732  x  1000  =  34673. 2 .    Ans.  34673. 2  M. 
Rule. — Multiply  the  given  quantity  hy  the  number  of  the 

required  denomination  which  mahes  a  unit  of  the  given 
denomination. 

Since  the  multiplier  is  always  10,  100,  1000,  etc.,  the  operation  is 
performed  by  removing  the  decimal  point  as  many  places  to  the  vigM 
as  there  are  ciphers  in  the  multiplier,  annexing  ciphers  when 
necessary. 

2.  Reduce  63.7504  hektograms  to  grams.  Ans,  6375.04  G. 

3.  Express  347.5  liektoliters  in  liters. 

4.  In  9.375  hektars,  liow  many  square  meters? 

5.  In  30.16  kilograms,  how  many  grams? 

6.  Change  1473  ars  to  square  meters. 

7.  Express  514  cubic  meters  in  cubic  decimeters. 

8.  Reduce  7.63  kilos  to  grams. 

•sLOS.  To  express  a  lower  Metric  Denomination  in 
terms  of  its  higher  denominations. 

1.  Reduce  1073.56  meters  to  kilometers. 
Operation.     1073.56-^1000  =  1.07356.    Aub.  1.07356  Km. 
Rule. — Divide  the  given  quantity  hy  the  number  of  its 

own  denomination  lohich  makes  a  unit  of  the  required 
denomination. 

Since  the  divisor  is  always  10,  100,  1000,  etc.,  the  operation  is 
performed  by  removing  the  decimal  point  as  many  places  to  the 
left  as  there  are  ciphers  in  the  divisor,  prefixing  ciphers  when 
necessary. 

2.  Reduce  40063  liters  to  hektoliters.  Ans.  400.63  HI. 

3.  In  136.7  centigrams,  how  many  grams  ?  Ans.  1.367  G. 

4.  In  456021  square  meters,  how  many  hektars  ? 

Ans.  45.6021  Ha. 

5.  Express  512.63  grams  in  kilograms. 

6.  How  many  cubic  meters  or  stars  in  76341  cu.  deci- 
meters ? 

7.  In  52000  centimeters,  how  many  dekameters  ? 


352  METRIC     SYSTEM. 

494.  To  add,  subtract,  multiply,  and  divide  Metric 
D  enominations. 

1.  What  is  the  sum  of  314.217  liters,  53.062  hektoliters, 
and  125  milliliters  ? 

OPEBATioy.    314.217  L.  +  5306.3  L.  +  .125  L.  =  5620.543  L.,  Ane, 

2.  Find  the  difference  between  4.37  kilometers,  and  1242 
meters. 

Opekation.    4.37  Kiii.-1.242  Km.  =  3.128  Km.,  Am. 

3.  How  much  cloth  in  SJ  pieces  of  cloth,  each  piece  con- 
taining 43.65  meters  ? 

Operation.    43.65  M.  x  8.25  =  360.1125  M.,  Ans, 

4.  How  many  vessels,  each  containing  3.5  liters,  can  be 
filled  from  43.75  dekaliters  of  wine? 

Operation.    437.5  L.^3.5  L.  =  135 ;  125  vessels,  Ans. 

KuLE.  Write  the  given  numbers  decimally,  reducing  to 
the  same  denomination  when  necessary,  and  then  proceed 
as  in  the  corresponding  operations  of  whole  numbers  and 
decimals, 

5.  Find  the  sum  of  256.4  L.,  7.641  HI.,  and  843  Kl. 

6.  From  25.8  G.  take  326  centigrams.  How  much  re- 
mains ? 

7.  A  dealer  bought  15  tons  of  coal,  and  sold  at  one 
time  5.32  T.,  at  another  3.045  T.,  and  at  another  4.125  T. 
How  much  had  he  left  ? 

8.  From  a  piece  of  cloth  containing  45.75  M.,  a  tailor  cut 
5  suits  of  clothes,  each  suit  containing  7.5  M.  How  much 
remained  ? 

9.  What  is  the  difference  between  2  Ha.  and  1242  ca.  ? 

10.  Find  the  sum  ofl2.65  M.,  3.678  Dm.,  2106.32  Hm., 
and  570.36  dm. 

11.  Find  the  difference  between  157  S.  and  1.6  Ds. 


METRIC    SYSTEM.  353 

495.  To  change  the  Metric  to  the  Common 
System. 

I.  3.6  Km.,  how  many  feet  ? 

OPERATION.  Analysis.     The  meter  is 

3.6  Km.  X  1000  =  3600  M.  the  principal  unit  of  the  table ; 

39.37  in.  x  3600  =  141732  in,       ^^^^  ^®  ^^^^^^e  the  kilome- 

1  4ir^nc\  '  -in  •  110-1-1  ^i.      ters  to  meters.     Since  there 

141732  m.-^  12  in.= 11811  ft.  _._.    ,      .  ^     . 

are  39.37  inches  m  a  meter,  in 

3600  meters  there  are  3600  times  39.37  inches  ;  and  since  there  are 

12  inches  in  a  foot,  there  are  as  many  feet  as  12  inches  are  contained 

times  in  141732  inches.    Therefore  3.6  Km.  are  equal  to  11811  feet. 

EuLE.  Reduce  the  given  quantity  to  the  denomination  of 
the  principal  unit  of  the  table  ;  multiply  hy  the  equivalent^ 
and  reduce  the  product  to  the  required  denomination. 

Examples  foe  Peactice. 

2.  How  many  feet  in  472  centimeters  ? 

Ans.  15.4855^  ft. 

3.  How  many  cubic  feet  in  2000  sters  ? 

Ans.  70632  cu.  ft. 
4.. How  many  gallons,  wine  measure,  in  325  deciliters? 

Ans.  8  gals.  2.343— qts. 
6.  How  many  gallons  in  108.24  liters  ? 

Ans,  28.594+ gals. 

6.  How  many  bushels  in  3262  kiloliters  ? 

Ans.  92559.25  busli. 

7.  How  many  square  yards  in  436  ars  ? 

Ans.  52145.6  sq.  yds. 

8.  In  942325  centiliters,  how  many  bushels  ? 

Ans.  267.3847 +bush. 

9.  In  436  kilograms,  how  many  pounds  ? 

Ans.  961.19314  lbs.  ' 
10.  In  42  ars,  how  many  square  rods  ? 

II.  75.5  hektars  to  acres  ? 

12.  In  450  cubic  meters,  how  many  cubic  feet  ? 

13.  How  many  gallons  in  24J  liters  of  v/ine  ? 

14.  IIow  many  sters  in  18  cords  of  wood  ? 

15.  How  many  pounds  of  butter  in  84  kilos  ? 


354  METRIC     SYSTEM. 

496.  To  change  from  the  Common  to  the  Metric 
System. 

1.  In  10  lbs.  4  oz.  how  many  kilograms  ? 

OPERATION.  Analysis.     The 

10  lbs.  4  oz.  =  10.25  lbs.  gram,  the  principal 

10.25  lbs.  X  7000  =  71750  gr.  ^^i*  of  the  table,  is 

71750  gr.-^15.432  gr.  =  4649.43  G.  expressed  in  grains; 

4649.43  G.-MOOO  =  4.64943  Kg.,  Ans.  ^"^^ 

to  grains.  15.432  grains  make  one  gram  ;  hence  there  are  as  many- 
grams  in  71750  grains  as  15.432  grains  is  contained  times  in  71750 
grains.  And  since  there  are  1000  grams  in  a  kilogram,  dividing 
4649.43  G.  by  1000  will  give  the  kilograms  in  10  lb.  4  oz.  Therefore, 
10  lb.  4  oz.  is  equal  to  4.64943  Kg. 

EuLE.  Reduce  the  given  quantity  to  the  denomination  in 
which  the  equivalent  of  the  principal  unit  of  the  metric  table 
is  expressed  j  divide  ly  this  equivalent,  and  reduce  the  quo- 
tient to  the  required  denomination* 

Examples  foe  Practice. 

2.  In  6172.8  pounds,  how  many  kilograms  ? 

Ans.  2800  Kg. 

3.  How  many  hektars  in  2392  square  yards  ? 

Ans.  .2  Ha. 

4.  How  many  ars  in  a  square  mile  ? 

Ans,  25899.665552  A. 

5.  How  many  cubic  dm.  in  18924  cubic  yards  ? 

Ans.  14467889.9082  cu.  dm. 

6.  In  892  grains,  how  many  grams  ? 

Ans.  57.8019  G. 

7.  In  2  miles,  279  rods,  and  5  yards,  how  many  kilo- 
meters? Ans.  4.626418  Km. 

8.  Bought  454  bush,  wheat  at  $3,  and  sold  the  same  at 
$8.75  per  hektoliter ;  how  many  hektoliters  did  I  sell  ?  Did 
I  gain,  or  lose,  and  how  much  ? 

Ans.  160  HI.     Gain,  $38. 


METRIC    SYSTEM.  355 

497.    MISCELLANEOUS    EXAMPLES. 

Eequired,  the  footings  of  the  following  bills  : 

New  York,  April  23, 1877. 


W.  J.  Milne, 

BoH  of  L.  CooLEY  &  Son. 

122    M  Broadcloth, 

@    16.00 

320     ''   Bid.  Shirting, 
230     ''  White  Flannel, 

.35 
.30 

206.5  ''  Ticking, 
107.9  ''   Blk.  Silk, 

.31 
''      2.40 

Rec'd  Payment, 

Ans.              $1235.975 
L.  CooLEY  &  Son. 

(2 
Chas.  D.  McLean, 

Buffalo,  May  1,  1878. 

BoH  of  Wm.  Benedict. 
40  chests  Tea,  each  30.5  Kg.  @  I  2.50 
12  sacks  Java  Coffee,  ''  40.00 
25  bbls.  Coffee  Sugar,  each  110  Kg.  ''  .32 
10  "  Crushed  ''  ''  95  Kg.  ''  .38 
30  boxes  Raisins,  "      12  Kg.  ''        .50 

Ans,  $4951.00 

Redd  Payment, 

Wm.  Benedict. 

3.  A  man  bought  a  lot  of  land  4  Km.  long  and  2  Km. 
wide,  and  sold  one-third  of  it.  How  many  ars  had  he  left, 
and  what  was  the  cost  of  the  lot,  at  $100  per  acre  ? 

Ans.  to  first,  53333.33^  A.    Ans.  to  second,  $197685.95. 

4.  A  farmer  sold  540  HI.  of  wheat  at  $6,  and  invested  the 
proceeds  in  coal  at  $8  per  ton.  How  many  metric  tons  of 
coal  did  he  purchase  ?  Ans.  367.41835147  tons. 

5.  What  will  be  the  cost  of  a  pile  of  wood  42.5  M.  long, 
2  M.  high,  1.9  M.  wide,  at  $2  per  ster?  Ans.  $323. 


35G  METRIC     SYSTEM. 

6.  How  many  meters  of  shirting,  at  $.25  par  meter,  must 
be  given  in  exchange  for  300  HI.  oats,  at  $1.20  per  hekto- 
liter  ?  Ans,  1440  M. 

7.  A  grocer  buys  butter  at  $.28  per  lb.,  and  sells  it  at  $.60 
per  kilogram.    Does  he  gain  or  lose,  and  what  per  cent  ? 

Ans,  Lost,  2-|^  %, 

8.  A  bin  of  wheat  measures  5  M.  square,  and  2.5  M.  deep. 
How  many  hektoliters  will  it  contain,  and  what  will  be  the 
cost  of  the  wheat,  at  $2  per  bushel  ? 

Ans.  625  HI.     $3546.875. 

9.  What  price  per  pound  is  equivalent  to  $2.50  per  Hg.  ? 

Ans.  $11.34. 

10.  A  merchant  bought  240  M.  of  silk  at  $2,  and  sold  it 
at  $1.95  per  yard.     Did  he  gain  or  lose,  and  how  much? 

Ans.  Gain,  $31.81. 

11.  What  is  the  area  of  a  floor  1.25  M.  long,  and  8.7  M. 
wide  ?  Ans.  .10875  A. 

12.  A  merchant  shipped  to  France  50  bbl.  of  coffee  sugar, 
each  containing  250  lb.,  paying  $2  per  hundred  for  trans- 
portation. He  sold  the  sugar  at  $.34  per  kilogram,  and  in- 
vested the  proceeds  in  broadcloth  at  $4  per  meter.  How 
many  yards  of  broadcloth  did  he  purchase  ? 

Ans.  458.71 4- yds. 

13.  What  would  be  the  cost  of  a  pile  of  wood  15.7  M.  long, 
3  M.  high,  and  7.52  M.  wide,  at  $1.50  a  ster  ? 

Ans.  $531.29. 

14.  What  would  be  the  cost  of  excavating  a  cellar  18.3  M. 
long,  10.73  M.  wide,  and  3.4  M.  deep,  at  15  cents  per  ster  ? 

Ans.  $100.14. 

15.  How  many  centars  of  plastering  in  a  house  contain- 
ing six  rooms  of  the  following  dimensions,  deducting  one- 
twelfth  for  doors,  windows,  and  base  ?  and  what  would  be 
the  cost  of  the  work  at  38  cents  per  centar  ?  First  room, 
6.2  M.  X  4.7  M.;  second  room,  4.52  M.  x  4  M. ;  third  room, 

6  M,  X  5.2  M. ;  fourth  room,  3.82  M.  x  3.82  M.;  fifth  room, 

7  M.X6.2  M.;   sixth  room,  4.5  M.  x4.25  M.     Height  of 
each  room,  3.8  M.  Ans.  562.030  ca.     $213.57. 


MISCELLANEOUS.  357 


MISCELLANEOUS  TABLES. 

498.  The  old  French  Linear,  and  Land  Measure, 

is  still  used  to  some  extent  in  Louisiana,  and  in  other 
French  settlements  in  the  United  States. 


Table. 

12  Lines    =  1  Inch.  6  Feet     =  1  Toise. 

12  Inches  =  1  Foot.  32  Toises  =  1  Arpent. 

900  Square  Toises  =  1  Square  Arpent. 

The  FreThch  Foot  equals  12.8  inches,  American,  nearly. 
The  Ar'perd  is  the  old  French  name  for  Acre^  and  contains  nearly 
\  of  an  English  acre. 

In  Texas,  New  Mexico,  and  in  other  Spanish  settle- 
ments of  the  United  States,  the  following  denominations 
are  still  used : 


Table. 

1000000  Square  Varas  =  1  Labor    =    177.136  Acres  (American). 
25  Labors  =  1  League  =  4428.4     Acres  " 

The  Spanish  Foot  =  11.11+  in.  (Am.) ;  1  Vara  =  33|  in.  (Am.); 
108  Varas  =  100  Yards,  and  1900.8  Varas  =  1  Mile. 

Other  Denominations  m  Use. 

6000        Varas  Square  =       1  Square  League. 
1000        Varas  Square  =        1  Labor,  or  ^  League. 
5645.376  Square  Varas  =  4840  Square  Yards  =      1  Acre. 
23.76    Square  Varas  =        1  Sqi/are  Chain  =    y*^  Acre. 
1900.8     Varas  Square  =       1  Section  =  640  Acres. 


^58 


MISCELLANEOUS. 


499.  The  following  table  will  assist  farmers  in  making 
\n  accurate  estimate  of  the  amount  of  land  in  different 
Selds  under  cuMvation  : 


Table. 


10  rods  X     16  rods  -  1 

L  A. 

220  feet 

X 

198 

8     "      X     20     "     ^  3 

110     " 

X 

369 

6     "      X     33     «     -  ] 

60     " 

X 

726 

4     "      X     40     «     =  ] 

120     " 

X 

303 

5  yds.    X   DSS  ^ds^    -  1 

200     '* 

X 

108.9 

10     "      X   484     '^      =  ] 

100     " 

X 

145.2 

26     *•      X   24£     "'      i=  : 

100     '* 

X 

108.9 

40     "      X   12i     ''      t  ] 

feet  = 


500.  The  foU'^inng  table  will  often  be  found  conven- 
ient, taking  i^yside  dimensions : 

A  box  24  in.  x  24  in.  x  14.7  will  contain  a  harrd  of  81 J  gallons. 

A  box  15  in.  x  14  in.  x  11  in.  will  contain  10  gaUons, 

A  box  8^  in.  x  7  in.  x  4  in.  will  contain  a  gallon. 

A  box  4  in.  x  4  in.  x  3.6  in.  will  contain  a  quart. 

A  box  24  in.  x  28  in.  x  16  in.  will  contain  5  bushels, 

A  box  16  in.  x  12  in.  x  11.2  in.  will  contain  a  hushd, 

A  box  12  in.  x  11.2  in.  x  8  in.  will  contain  a  Tialf-bushel. 

A  box  7  in.  x  6.4  in.  x  12  in.  will  contain  a  peck. 

A  box  8. 4  in.  X  8  in.  X  4  in.  will  contain  a  half -peck  or  4  dry  quarts. 

A  box  6  in,  by  5 1  in.,  and  4  in.  deep,  will  contain  a  half -gallon. 

A  box  4  in.  by  4  in.,  and  2^^  in.  deep,  will  contain  a  pint. 

501.  Nails  are  put  up  100  pounds  to  the  keg. 


SiZB. 

ii 

1' 

SiZB. 

=3c 

SiZB. 

is 

^.a 

^B 

^5 

►3.9 
2 

^.g 

dd  fine  blued. 

u 

725 

dOd  com.  blued. 

H 

16 

6d  casing. 

210 

Sd  com.     " 

H 

400 

40d    " 

5 

14 

8d     " 

2| 

134 

4d    *' 

u 

300 

50^    "        " 

5^ 

11 

lOd     *' 

3 

78 

6d    " 

2 

150 

60d    " 

6 

8 

Qd  finishing 

2 

317 

Sd    **        " 

21 

85' 

Qd  fence. 

2 

80 

8d     " 

2\ 

208 

IM  " 

3 

60 

8d      " 

3| 

50 

16d     '•■ 

3 

126 

\M  " 

n 

50 

lOd      " 

3 

30 

6d  clinching 

2 

118 

16d  « 

40 

12d      '• 

II 

27 

8d     " 

2| 

80 

20d  « 

? 

30l 

IQd      " 

20 

lOd     " 

3 

45 

5  lbs.  of  4rf  or  3J  lbs.  of  Sd  will  put  on  1,000  shingles. 
5J  lbs.  of  3d  fine  will  put  on  1 ,000  lath. 


MISCELLANEOUS 


359 


Raileoad  Freight. 

502.  When  conyenient  to  weigh  them,  all  goods  are 
billed  at  actual  weight ;  but  ordinarily,  the  articles  named 
below  are  billed,  at  the  rates  given  in  the  following 

Table. 


Ale  or  Beer, 

820  lbs 

per  bbl. 

Highwines,        350  lbs.  per  bbl 

Apples,  green, 

150  *' 

<< 

Lime,                 200  " 

Beef. 

320  " 

" 

Nails,                  108  "    per  keg. 

Barley, 

48  " 

per  bu. 

Oil,                     400  «     perbbL 

Beans, 

60  " 

" 

Oats,                    32  "     perbu. 

Cider, 

350  - 

per  bbl. 

Pork,                 320  "     perbbL 

Corn  Meal, 

220  " 

" 

Potatoes,  com'n,150  "            *' 

Com,  shelled. 

56  " 

per  bu. 

Salt,  fine,           300  " 

Corn  in  ear, 

70  " 

€( 

"     coarse,      350   " 

Clover  Seed, 

60  '* 

« 

''    in  sacks,  200  "   per  sack. 

It 

200  " 

per  bbl. 

Wheat,                60  «   perbu. 

300  " 

n 

Whiskey,          350  "   perbbL 

Flour, 

200  " 

it 

2000  pounds  are  reckoned  1  Toti. 

Generally  from  18000  to  20000  pounds  is  considered  a  car  load. 


503.  Lumber  and  some  other  articles  are  estimated  as 
follows : 


Weight. 
Pine,  Hemlock,  and  Poplab,  thoroughly 

seasoned,  per  thousand  feet 3000 

Black  Walnut,  Ash,  Maple,  and  Chekby, 

per  thousand  feet 4000 

Pine,  Hemlock,  and  Poplar,  green,  per  M .  4000 
Black  Walnut,  Ash,  Maple,  and  Cherry, 

green,  per  M 4500 

Oak,  Hickory,  and  Elm,  dry,  per  M 4000 

Oak,  Hickory,  and  Elm,  green,  per  M 5000 

Shingles,  green,  per  thousand 375 

Lath,  per  thousand 500 

Brick,  common,  per  car  load 4  lbs  each* 

Coal,  per  car  load 

Stone,  undressed,  per  cubic  yard 4000 


Amount  for 
car  load. 


6500 

5000 
5000 

4000 
5000 
4000 

35  M. 

40  M. 
5000 
250  bu. 
5cu.yd. 


Table  for  Iitvestors. 

504.  Thefollmmng  Table  shows  the  rate  per  cent,  of  Annual  Income 
from  Bonds  hearing  5,  6,  7,  or  8  per  cent,  interest,  and  costing 
from  40  to  125. 


I*urchase 
Price. 

5%. 

6^. 

7/.. 

8/.. 

Purchase 
Price. 

5f.. 

6/.. 

nfc. 

8/.. 

40 

12.50 

15.00 

17.50 

20.00 

83 

6.02 

7.22 

8.43 

9.63 

41 

12.20 

14.64 

17.08 

19.52 

84 

5.95 

7.14 

8.33 

9.52 

42 

11.90 

14.28 

16.66 

19.04 

85 

5.88 

7.05 

8.23 

9.41 

43 

11.63 

13.95 

16.28 

18.61 

86 

5.81 

6.97 

8.13 

9.30 

44 

11.36 

13.63 

15.90 

18.18 

87 

5.74 

6.89 

8.04 

919 

45 

11.11 

13.32 

15.56 

17.78 

88 

5.68 

6.81 

7.94 

9.09 

46 

10.86 

13.04 

15.21 

17.39 

89 

5.61 

6.74 

7.86 

8.98 

47 

10.63 

12.77 

14.90 

17.02 

90 

5.55 

6.66 

7.77 

8.88 

48 

10.41 

12.50 

14.53 

10.66 

91 

5.49 

6.59 

7.69 

8.79 

49 

10.20 

12.25 

14.29 

16.33 

92 

5.43 

6.52 

7.60 

8.69 

50 

10.00 

12.00 

14.00 

16.00 

93 

5.37 

6.45 

7.52 

8.60 

51 

9.80 

11.76 

13.72 

15.68 

94 

5.31 

6.38 

7.44 

8.51 

52 

9.61 

11.53 

13.40 

15.38 

95 

5.26 

6.31 

7.36 

8.43 

53 

9.43 

11.32 

13.20 

15.09 

96 

5.20 

6.25 

7.29 

8.33 

54 

9.25 

11.11 

12.96 

15.81 

97 

5.15 

6.18 

7.21 

8.24 

55 

9.09 

10.90 

12.72 

14.54 

98 

5.10 

6.12 

7.14 

8.16 

56 

8.92 

10.70 

12.50 

14.28 

99 

5.05 

6.06 

7.07 

8.08 

57 

8.77 

10.52 

12.27 

14.03 

100 

5.00 

6.00 

7.00 

8.00 

58 

8.62 

10.34 

12.06 

13.79 

101 

4.95 

5.94 

6.93 

7.93 

59 

8.47 

10.18 

11.86 

13.55 

102 

4.90 

5.88 

6.86 

7.84 

CO 

8.33 

10.00 

11.66 

13  33 

103 

4.85 

5.82 

6.79 

7.76 

CI 

8.19 

9.83 

11.47 

13.11 

104 

4.80 

5.76 

6.72 

7.69 

62 

8.06 

9.07 

11.29 

12.90 

105 

4.76 

5.71 

666 

7.61 

C3 

7.93 

9.52 

11.11 

12.69 

106 

4.71 

5.66 

6.60 

7.54 

64 

7.81 

9.37 

10.93 

12.50 

107 

4.67 

5.60 

6.54 

7.47 

65 

7.69 

9.23 

10.76 

12.30 

108 

4.62 

5.55 

6.48 

7.40 

66 

7.57 

9.09 

10.60 

12.12 

109 

4.58 

5.50 

6.42 

7.33 

67 

7.43 

8.95 

10.44 

11.94 

110 

4.54 

5.45 

6.36 

727 

68 

7.35 

8.82 

10.29 

11.76 

111 

4.50 

5.40 

6.30 

7.20 

69 

7.24 

8.69 

10.14 

11.50 

112 

4.46 

5.35 

6.25 

7.14 

70 

7.14 

8.57 

10.00 

11.43 

113 

4.42 

5.30 

6.19 

7.07 

71 

7.04 

8.45 

9.85 

11.26 

114 

4.38 

5.26 

6.14 

7.01 

72 

6.94 

8.33 

9.72 

11.11 

115 

4.35 

5.21 

6.08 

6.95 

73 

6.84 

8.21 

9.58 

10.95 

116 

4.31 

5.17 

6.03 

6.89 

74 

6.75 

8.10 

9.45 

10.80 

117 

4.27 

5.13 

5.98 

6.83 

75 

6.66 

8.00 

9.33 

10.66 

118 

4.23 

5.08 

5.93 

6.77 

76 

6.57 

7.89 

9.21 

10.52 

119 

4.20 

5.04 

5.88 

6.73 

77 

6.49 

7.79 

9.00 

10.38 

120 

4.16 

5.00 

5.83 

6.66 

78 

6.41 

7.69 

8.97 

10.25 

121 

4.13 

4.95 

5.78 

6.61 

79 

6.32 

7.59 

8.86 

10.12 

123 

4.09 

4.91 

5.73 

6.55 

80 

6.25 

7.50 

8.75 

10.00 

123 

4.06 

4.87 

5.69 

6.50 

81 

6.17 

7.40 

8.64 

9.87 

124 

4.03 

483 

5.65 

6.45 

83 

6.09 

7.31 

8.53 

9.75 

125 

4.00  1  4.80  5.60 

6.40 

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i 


14  DAY  USE 

RETURN  TO  DESK  FROM  WHICH  BORROWED 

EDUCATION-PSYCHOLOGY 
LIBRARY 

This  book  is  due  on  the  last  date  stamped  below,  or 

on  the  date  to  which  renewed. 

Renewed  books  are  subject  to  immediate  recalL 


IIM   9     1969 

JUL  9    REC'D-8 

AM 

"^ikfAAx-ar         "--i^g^"^" 

rti  35838 


m 


Xotable   Jfeiv    Boohs. 
SWINTOK'S 

Supplementary  Readers. 


A 


ED.iED  BY 

WILLIAM  SWINTON  and  QEORaE  R.  CATHCAKT. 

SERIES  OF  CAREFULLY  GRADI^ATED  READING  BOOKS, 
designed  to  connect  with  any  of  the  regular  Readers.  They  are  attrac- 
tive In  appearance,  are  bound  in  cloth,  and  the  iirst  tour  books  are  pro- 
fusely illustratca  by  the  best  artists.  The  six  boo'  •  which  are  closely  co-ordi- 
nated with  the  several  Reade/s  of  the  regular  series,  are: 

I.  EASY  STEPS  FOR  LITTLE  FEET. 

Svpplementnry  fo  First  Reader. 
Lt  this  book  the  attractive  is  the  chief  aim,  and  the  pieces  have  been  written 
and  chcsen  with  special  reterence  to  the  feelings  and  fancies  of  early  childhood. 

II.   GOLDEN  BOOK  OF  CHOICE  READING. 

Sufplementary  to  Second  Reader. 
This  book  represents  a  grt. at  variety  of  pleasing  and  instructive  rjadmg, 
C'-nSstJng  of  child-lore  ?.nd 'poetry,  noble   examples  and  attractive  object- 
ri»ading,  vriiten  specially  ior  it. 


Ttt;  book  of  TALEJ. 

•leanings  Jvzaginathe  and  Emotional.      SvJ'plcutenta^ 
Thikd  Reader. 


ftx  this  u;-.,k  the  youthful  taste  f  r  imaginptive  and  emotional  is  fed  wuh 
purf,  and  noble  creations  drawn  Irom  the  literature  of  all  nations. 

IV.  READINGS  IN  NATU^^'S  BOOK. 

Si<J>f>ieji!ft}tarytoVoVKTv\i^L'J.  -KR. 
Tir's  book  contains  a  varied  collection  of  charming  rondings  in  Natural  His- 
tory and  Botany,  drawn  from  the  works  of  the  great  modern  naturalists  and 
travelers. 

J.  SEVEN  AMERICAN  CLASS  C3. 
VI.  SEVEN  BRITISH  CLASSICS.    > 

The  "Classics"  arc  suitable  fo.  reading  in  advanced  grades,  and  aim  10 
instill  a  tasic  lor  the  highev  literature,  by  tiie  prcsentauon  of  get^isof  Dritish 
and  American  authorship. 


■r 


.  A  SHORTER  COQRSE  IN 

htGlim  GRAMMAR  AND  COIPOSITIOIJ, 


By  W.  IT.  WELLS,  A.  M. 


,A.n  >i  tire! y  new  work  by  this  venerable  author,  biMu 
long  study  of  books  and  rnethuiis  in  this  branch  of  educati  >i.. 

"Tio  work  is  intended  to- furiush  all  me  aid  that  is  nr    '    '  '   -  ■' 
student  p  learning  to  spcr.k  and  writo  utW.    It  begl' 
vvl.ioh  the;  pupil  already  pnv'^r  so^,  and  carries  Wm  dir."  ' 
leal  study  and  use  of  \h  c.      The  rules  oi 

taught  A^  ah'itract  princii  :  y  where  in  their  pi 

This  Shorter  Cucbsl  '  ^V  be  defined  as  a  1 

ercises  in  speaking  and  wruini-  r  iglish,  accompanictl  bv  >,  nj,.  ,. 
tion  of  the  Principles  and  Rules  b  v  which  language  is  guv'i.-riicd. 

*<«  Cotrcsfondence   is  soliettcU  xoif/i   r^ferenc  to  the  xt:trod:n. 

abovt  Ifiol-t,    fuvoralie  rates  -will  be  made  fof  ivannn-ition  and  in 

JVISON,  BIAKI-MAN,  L\Y!0I{7 

TJT^.A'.''  YOr.K  -lucl  CHICAGO. 


the  result  of  a  life- 


yiiH 


